#set document(title: "5.1 Inverse Functions", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 5.1#h(0.6em)Inverse Functions === Introduction When you buy a house, your monthly mortgage payment is a function of the size of the loan. The table shows mortgage payments on #math.equation(block: false, alt: "30")[$30$]-year loans of various sizes at #math.equation(block: false, alt: "6")[$6$]% interest. #figure(table( columns: 6, align: left, inset: 6pt, table.header([Loan amount, #math.equation(block: false, alt: "L")[$L$]], [#math.equation(block: false, alt: "150 , 000")[$150 , 000$]], [#math.equation(block: false, alt: "175 , 000")[$175 , 000$]], [#math.equation(block: false, alt: "200 , 000")[$200 , 000$]], [#math.equation(block: false, alt: "225 , 000")[$225 , 000$]], [#math.equation(block: false, alt: "250 , 000")[$250 , 000$]]), [Mortgage payment, #math.equation(block: false, alt: "M")[$M$]], [#math.equation(block: false, alt: "899.33")[$899.33$]], [#math.equation(block: false, alt: "1049.21")[$1049.21$]], [#math.equation(block: false, alt: "1199.10")[$1199.10$]], [#math.equation(block: false, alt: "1348.99")[$1348.99$]], [#math.equation(block: false, alt: "1498.88")[$1498.88$]], )) For the function #math.equation(block: false, alt: "M equals f open parenthesis L close parenthesis")[$M = f ( L )$], the input value is the amount of the loan, and the output is the mortgage payment. However, when you are shopping for a house, you may think of the mortgage payment as the input variable: If you can afford a certain monthly mortgage payment, how large a loan can you finance? Now the mortgage payment is the input value, and the loan amount is the output. By interchanging the inputs and outputs, we define a new function, #math.equation(block: false, alt: "L equals g open parenthesis M close parenthesis")[$L = g ( M )$], shown below. #figure(table( columns: 6, align: left, inset: 6pt, table.header([Mortgage payment, #math.equation(block: false, alt: "M")[$M$]], [#math.equation(block: false, alt: "899.33")[$899.33$]], [#math.equation(block: false, alt: "1049.21")[$1049.21$]], [#math.equation(block: false, alt: "1199.10")[$1199.10$]], [#math.equation(block: false, alt: "1348.99")[$1348.99$]], [#math.equation(block: false, alt: "1498.88")[$1498.88$]]), [Loan amount, #math.equation(block: false, alt: "L")[$L$]], [#math.equation(block: false, alt: "150 , 000")[$150 , 000$]], [#math.equation(block: false, alt: "175 , 000")[$175 , 000$]], [#math.equation(block: false, alt: "200 , 000")[$200 , 000$]], [#math.equation(block: false, alt: "225 , 000")[$225 , 000$]], [#math.equation(block: false, alt: "250 , 000")[$250 , 000$]], )) This new function gives the same information as the original function, #math.equation(block: false, alt: "f")[$f$], but from a different point of view. We call the function #math.equation(block: false, alt: "g")[$g$] the #strong[inverse function] for #math.equation(block: false, alt: "f")[$f$]. The elements of the #emph[range] of #math.equation(block: false, alt: "f")[$f$] are used as the input values for #math.equation(block: false, alt: "g")[$g$], and the output values of #math.equation(block: false, alt: "g")[$g$] are the corresponding domain elements of #math.equation(block: false, alt: "f")[$f$]. For example, from the tables you can verify that #math.equation(block: false, alt: "f open parenthesis 200 , 000 close parenthesis equals 1199.10")[$f ( 200 , 000 ) = 1199.10$], and #math.equation(block: false, alt: "g open parenthesis 1199.10 close parenthesis equals 200 , 000")[$g ( 1199.10 ) = 200 , 000$]. In fact, this property defines the inverse function. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Functions] Suppose #math.equation(block: false, alt: "g")[$g$] is the #strong[inverse function] for #math.equation(block: false, alt: "f")[$f$]. Then #math.equation(block: true, alt: "g open parenthesis b close parenthesis equals a if and only if f open parenthesis a close parenthesis equals b")[$g ( b ) = a " " " " " " " " " " " if and only if " " " " " " " " " " " f ( a ) = b$] ] #examplebox("Example 1")[][ Suppose #math.equation(block: false, alt: "g")[$g$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], and we know the following function values for #math.equation(block: false, alt: "f")[$f$]: #math.equation(block: true, alt: "f open parenthesis minus 3 close parenthesis equals 5 , f open parenthesis 2 close parenthesis equals 1 , f open parenthesis 5 close parenthesis equals 0")[$f ( − 3 ) = 5 , " " " " f ( 2 ) = 1 , " " " " f ( 5 ) = 0$] Find #math.equation(block: false, alt: "g open parenthesis 5 close parenthesis")[$g ( 5 )$] and #math.equation(block: false, alt: "g open parenthesis 0 close parenthesis")[$g ( 0 )$]. #solutionbox[ We know that #math.equation(block: false, alt: "g open parenthesis 5 close parenthesis equals minus 3")[$g ( 5 ) = − 3$] because #math.equation(block: false, alt: "f open parenthesis minus 3 close parenthesis equals 5")[$f ( − 3 ) = 5$], and #math.equation(block: false, alt: "g open parenthesis 0 close parenthesis equals 5")[$g ( 0 ) = 5$] because #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis equals 0")[$f ( 5 ) = 0$]. Tables may be helpful in visualizing the two functions, as shown below. #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]], [], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "0")[$0$]], )) #math.equation(block: true, alt: "→ Interchange the columns →")[$" " " " " " " " " " " " "→ Interchange the columns →"$] #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis")[$x = g ( y )$]], [], [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], )) For the function #math.equation(block: false, alt: "f")[$f$], the input variable is #math.equation(block: false, alt: "x")[$x$] and the output variable is #math.equation(block: false, alt: "y")[$y$]. For the inverse function #math.equation(block: false, alt: "g")[$g$], the roles of the variables are interchanged: #math.equation(block: false, alt: "y")[$y$] is now the input and #math.equation(block: false, alt: "x")[$x$] is the output. ] ] Suppose #math.equation(block: false, alt: "g")[$g$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], and suppose we know the following function values for f : #math.equation(block: true, alt: "f open parenthesis minus 1 close parenthesis equals 0 , f open parenthesis 0 close parenthesis equals 1 , f open parenthesis 1 close parenthesis equals 2")[$f ( − 1 ) = 0 , " " f ( 0 ) = 1 , " " f ( 1 ) = 2$] Find #math.equation(block: false, alt: "g open parenthesis 0 close parenthesis")[$g ( 0 )$] and #math.equation(block: false, alt: "g open parenthesis 1 close parenthesis")[$g ( 1 )$]. #math.equation(block: false, alt: "g open parenthesis 0 close parenthesis equals")[$g ( 0 ) =$]\_\_\_\_\_ #math.equation(block: false, alt: "g open parenthesis 1 close parenthesis equals")[$g ( 1 ) =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "g open parenthesis 0 close parenthesis equals minus 1")[$g ( 0 ) = − 1$]; #math.equation(block: true, alt: "g open parenthesis 1 close parenthesis equals 0")[$g ( 1 ) = 0$] ] Suppose #math.equation(block: false, alt: "g")[$g$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], and suppose we know the following function values for #math.equation(block: false, alt: "f")[$f$] : #math.equation(block: true, alt: "f open parenthesis minus 1 close parenthesis equals 0 , f open parenthesis 0 close parenthesis equals 1 , f open parenthesis 1 close parenthesis equals 2")[$f ( − 1 ) = 0 , " " f ( 0 ) = 1 , " " f ( 1 ) = 2$] Find #math.equation(block: false, alt: "g open parenthesis 0 close parenthesis")[$g ( 0 )$] and #math.equation(block: false, alt: "g open parenthesis 1 close parenthesis")[$g ( 1 )$]. #solutionbox[ #math.equation(block: true, alt: "g open parenthesis 0 close parenthesis equals minus 1")[$g ( 0 ) = − 1$]; #math.equation(block: true, alt: "g open parenthesis 1 close parenthesis equals 0")[$" " g ( 1 ) = 0$] ] If the point #math.equation(block: false, alt: "open parenthesis 3 , minus 4 close parenthesis")[$( 3 , − 4 )$] lies on the graph of #math.equation(block: false, alt: "f")[$f$], what point lies on the graph of its inverse function? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "open parenthesis minus 4 , 3 close parenthesis")[$( − 4 , 3 )$] ] If the point #math.equation(block: false, alt: "open parenthesis 3 , minus 4 close parenthesis")[$( 3 , − 4 )$] lies on the graph of #math.equation(block: false, alt: "f")[$f$], what point lies on the graph of its inverse function? + #math.equation(block: false, alt: "open parenthesis minus 3 , 4 close parenthesis")[$( − 3 , 4 )$] + #math.equation(block: false, alt: "open parenthesis minus 4 , 3 close parenthesis")[$( − 4 , 3 )$] + #math.equation(block: false, alt: "open parenthesis 4 , minus 3 close parenthesis")[$( 4 , − 3 )$] + #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 , minus the fraction 1 over 4 close parenthesis")[$( display(frac(1, 3)) , − display(frac(1, 4)) )$] === Finding a Formula for the Inverse Function If a function is given by a table of values, we can interchange the columns (or rows) of the table to obtain the inverse function. Swapping the columns works because we are really interchanging the input and output variables. If a function is defined by an equation, we can find a formula for its inverse function in the same way: Interchange the roles of the variables in the equation so that the old output variable becomes the new input variable. #examplebox("Example 2")[][ + The function #math.equation(block: false, alt: "H equals f open parenthesis t close parenthesis equals 6 plus 2 t")[$H = f ( t ) = 6 + 2 t$] gives the height of corn seedlings, in inches, #math.equation(block: false, alt: "t")[$t$] days after they are planted. Find a formula for the inverse function and explain its meaning in this context. + Make a table of values for #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and a table for its inverse function. #solutionbox[ + Write the equation for #math.equation(block: false, alt: "f")[$f$] in the form #math.equation(block: true, alt: "H equals 6 plus 2 t")[$H = 6 + 2 t$] In this equation, #math.equation(block: false, alt: "t")[$t$] is the input and #math.equation(block: false, alt: "H")[$H$] is the output. We interchange the roles of the variables by solving for #math.equation(block: false, alt: "t")[$t$] to obtain #math.equation(block: true, alt: "t equals the fraction H minus 6 over 2")[$t = frac(H − 6, 2)$] In this equation, #math.equation(block: false, alt: "H")[$H$] is the input and #math.equation(block: false, alt: "t")[$t$] is the output. The formula for the inverse function is #math.equation(block: true, alt: "t equals g open parenthesis H close parenthesis equals the fraction H minus 6 over 2")[$t = g ( H ) = display(frac(H − 6, 2))$] The function #math.equation(block: false, alt: "g")[$g$] gives the number of days it will take the corn seedlings to grow to a height of #math.equation(block: false, alt: "H")[$H$] inches. + To make a table for #math.equation(block: false, alt: "f")[$f$], we choose values for #math.equation(block: false, alt: "t")[$t$] and evaluate #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 6 plus 2 t")[$f ( t ) = 6 + 2 t$] at those #math.equation(block: false, alt: "t")[$t$]-values, as shown at left below.#figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "H equals f open parenthesis t close parenthesis")[$H = f ( t )$]], [], [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "H")[$H$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "12")[$12$]], )) #figure(table( columns: 2, align: left, inset: 6pt, [#math.equation(block: false, alt: "t equals g open parenthesis H close parenthesis")[$t = g ( H )$]], [], [#math.equation(block: false, alt: "H")[$H$]], [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "3")[$3$]], )) #linebreak() To make a table for #math.equation(block: false, alt: "g")[$g$], we could choose values for #math.equation(block: false, alt: "H")[$H$] and evaluate #math.equation(block: false, alt: "the fraction H minus 6 over 2")[$" " display(frac(H − 6, 2))$], but because #math.equation(block: false, alt: "g")[$g$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], we can simply interchange the columns in our table for #math.equation(block: false, alt: "f")[$f$], as shown at right above. #linebreak() You can check that the values in the second table do satisfy the formula for the inverse function, #math.equation(block: false, alt: "g open parenthesis H close parenthesis equals the fraction H minus 6 over 2")[$" " g ( H ) = display(frac(H − 6, 2))$]. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Note once again that the two tables show the same relationship between #math.equation(block: false, alt: "t")[$t$] and #math.equation(block: false, alt: "H")[$H$], but the roles of input and output have been interchanged. The function #math.equation(block: false, alt: "f")[$f$] tells us the height of the seedlings after #math.equation(block: false, alt: "t")[$t$] days, and #math.equation(block: false, alt: "g")[$g$] tells us how long it will take the seedlings to grow to height #math.equation(block: false, alt: "H")[$H$]. ] Which of these is the inverse of the function #math.equation(block: false, alt: "h equals 3 m minus 2")[$h = 3 m − 2$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "m equals the fraction h plus 2 over 3")[$m = display(frac(h + 2, 3))$] ] Which of these is the inverse of the function #math.equation(block: false, alt: "h equals 3 m minus 2")[$" " h = 3 m − 2$]? + #math.equation(block: false, alt: "m equals the fraction h plus 2 over 3")[$m = display(frac(h + 2, 3))$] + #math.equation(block: false, alt: "m equals 2 h minus 3")[$m = 2 h − 3$] + #math.equation(block: false, alt: "m equals the fraction h minus 3 over 2")[$m = display(frac(h − 3, 2))$] + #math.equation(block: false, alt: "m equals the fraction 1 over 3 h minus 2")[$m = display(frac(1, 3 h − 2))$] Carol can burn 600 calories per hour bicycling and 400 calories per hour swimming. She would like to lose 5 pounds, which is equivalent to 16,000 calories. + Write an equation relating the number of hours of cycling, #math.equation(block: false, alt: "x")[$x$], and the number of hours swimming, #math.equation(block: false, alt: "y")[$y$], that Carol must spend to lose 5 pounds. \[Do not use commas: For example, enter "10000" rather than "10,000". #linebreak() Answer: \_\_\_\_\_ + Write #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. #linebreak() #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals")[$y = f ( x ) =$]\_\_\_\_\_ #linebreak() What does #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis equals")[$f ( 10 ) =$]\_\_\_\_\_ tell you? #linebreak() \_\_\_\_\_ + Find the inverse function, #math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis")[$x = g ( y )$]. #linebreak() #math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis equals")[$x = g ( y ) =$]\_\_\_\_\_ #linebreak() What does #math.equation(block: false, alt: "g open parenthesis 10 close parenthesis equals")[$g ( 10 ) =$]\_\_\_\_\_ tell you? #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "600 x plus 400 y equals 16 , 000")[$600 x + 400 y = 16 , 000$] + #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals 40 minus 1.5 x")[$y = f ( x ) = 40 − 1.5 x$]; #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis equals 25")[$f ( 10 ) = 25$]; If Carol cycles for 10 hrs, she must swim for 25 hrs. + #math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis equals 26 the fraction 2 over 3 minus the fraction 2 over 3 y")[$x = g ( y ) = 26 frac(2, 3) − frac(2, 3) y$]; #math.equation(block: false, alt: "g open parenthesis 10 close parenthesis equals 20")[$g ( 10 ) = 20$]; If Carol swims for #math.equation(block: false, alt: "10")[$10$] hrs, she must cycle for #math.equation(block: false, alt: "20")[$20$] hrs. ] Carol can burn 600 calories per hour bicycling and 400 calories per hour swimming. She would like to lose 5 pounds, which is equivalent to 16,000 calories. + Write an equation relating the number of hours of cycling, #math.equation(block: false, alt: "x")[$x$], and the number of hours swimming, #math.equation(block: false, alt: "y")[$y$], that Carol must spend to lose 5 pounds. + Write #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$" " y = f ( x )$]. What does #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis")[$f ( 10 )$] tell you? + Find the inverse function, #math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis")[$" " x = g ( y )$]. What does #math.equation(block: false, alt: "g open parenthesis 10 close parenthesis")[$g ( 10 )$] tell you? #solutionbox[ + #math.equation(block: false, alt: "600 x plus 400 y equals 16 , 000")[$600 x + 400 y = 16 , 000$] + #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals 40 minus 1.5 x")[$y = f ( x ) = 40 − 1.5 x$]; #math.equation(block: false, alt: "f open parenthesis 10 close parenthesis equals 25")[$f ( 10 ) = 25$]; If Carol cycles for 10 hrs, she must swim for 25 hrs. + #math.equation(block: false, alt: "x equals g open parenthesis y close parenthesis equals 26 the fraction 2 over 3 minus the fraction 2 over 3 y")[$x = g ( y ) = 26 frac(2, 3) − frac(2, 3) y$]; #math.equation(block: false, alt: "g open parenthesis 10 close parenthesis equals 20")[$g ( 10 ) = 20$]; If Carol swims for #math.equation(block: false, alt: "10")[$10$] hrs, she must cycle for #math.equation(block: false, alt: "20")[$20$] hrs. ] === Inverse Function Notation If the inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function, we denote the inverse by the symbol #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$], read "#math.equation(block: false, alt: "f")[$f$] inverse." This notation makes it clear that the two functions are related in a special way. For example, the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 6 plus 2 t")[$" " f ( t ) = 6 + 2 t$] in Example has inverse function #math.equation(block: false, alt: "f to the power minus 1 open parenthesis H close parenthesis equals the fraction H minus 6 over 2")[$" " f^(− 1) ( H ) = display(frac(H − 6, 2))$]. #examplebox("Example 3")[][ If #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals x cubed plus 2 ,")[$" " " " y = f ( x ) = x^(3) + 2 , " " " "$] find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 10 close parenthesis")[$" " " " f^(− 1) ( 10 )$]. #solutionbox[ We first find the inverse function for #math.equation(block: false, alt: "y equals x cubed plus 2")[$y = x^(3) + 2$] by solving for #math.equation(block: false, alt: "x")[$x$]: #math.equation(block: true, alt: "x cubed, equals y minus 2, Substract 2 from both sides.; x, equals the cube root of y minus 2, Take cube roots.")[$x^(3) & = y − 2 & & "Substract 2 from both sides." \ x & = root(3, y − 2) & & "Take cube roots."$] The inverse function is #math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis equals the cube root of y minus 2")[$x = f^(− 1) ( y ) = root(3, y − 2)$]. Now we evaluate the inverse function at #math.equation(block: false, alt: "y equals 10")[$y = 10$]: #math.equation(block: true, alt: "f to the power minus 1 open parenthesis 10 close parenthesis equals the cube root of 10 minus 2 equals 2")[$f^(− 1) ( 10 ) = root(3, 10 − 2) = 2$] ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Although the same symbol, #math.equation(block: false, alt: "minus 1")[$− 1$], is used for both reciprocals and inverse functions, the two notions are #emph[not] equivalent. That is, the inverse of a given function is usually not the same as the reciprocal of that function. In Example, note that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis")[$f^(− 1) ( y )$] is not the same as the reciprocal of #math.equation(block: false, alt: "f open parenthesis y close parenthesis")[$f ( y )$], because #math.equation(block: true, alt: "the fraction 1 over f open parenthesis y close parenthesis equals the fraction 1 over y cubed plus 2 but f to the power minus 1 open parenthesis y close parenthesis equals the cube root of y minus 2")[$frac(1, f ( y )) = frac(1, y^(3) + 2) " " " " " but " " " " " f^(− 1) ( y ) = root(3, y − 2)$] To avoid confusion, we use the notation #math.equation(block: false, alt: "the fraction 1 over f")[$display(frac(1, f))$] to refer to the reciprocal of the function #math.equation(block: false, alt: "f")[$f$]. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In Example, you can check that #math.equation(block: false, alt: "f open parenthesis 2 close parenthesis equals 10")[$f ( 2 ) = 10$]. In fact, the two statements #math.equation(block: true, alt: "f to the power minus 1 open parenthesis 10 close parenthesis equals 2 and f open parenthesis 2 close parenthesis equals 10")[$f^(− 1) ( 10 ) = 2 " " " " " and " " " " " f ( 2 ) = 10$] are equivalent; they convey the same information. This fact is a restatement of our earlier observation about inverse functions, this time using inverse function notation. ] If #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis equals 2")[$f ( 5 ) = 2$], what is #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 2 close parenthesis")[$f^(− 1) ( 2 )$] equal to? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "5")[$5$] ] If #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis equals 2")[$f ( 5 ) = 2$], what is #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 2 close parenthesis")[$f^(− 1) ( 2 )$] equal to? + #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] + #math.equation(block: false, alt: "the fraction 1 over 5")[$display(frac(1, 5))$] + #math.equation(block: false, alt: "5")[$5$] + #math.equation(block: false, alt: "minus 5")[$− 5$] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Functions] Suppose the inverse of #math.equation(block: false, alt: "f")[$f$] is a function, denoted by #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. Then #math.equation(block: true, alt: "f to the power minus 1 open parenthesis y close parenthesis equals x if and only if f open parenthesis x close parenthesis equals y")[$f^(− 1) ( y ) = x " " " " " " " " " if and only if " " " " " " " " " f ( x ) = y$] ] + If #math.equation(block: false, alt: "z equals f open parenthesis w close parenthesis equals the fraction 1 over w plus 3")[$z = f ( w ) = display(frac(1, w + 3))$], find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals")[$f^(− 1) ( 1 ) =$]\_\_\_\_\_ + Write two equations about the value of #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$], one using #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] and one using #math.equation(block: false, alt: "f")[$f$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals")[$f^(− 1) ( 1 ) =$]\_\_\_\_\_, and #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis equals")[$f ( − 2 ) =$]\_\_\_\_\_ + Show that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$] is not equal to #math.equation(block: false, alt: "the fraction 1 over f open parenthesis 1 close parenthesis")[$display(frac(1, f ( 1 )))$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals")[$f^(− 1) ( 1 ) =$]\_\_\_\_\_, but #math.equation(block: false, alt: "the fraction 1 over f open parenthesis 1 close parenthesis equals")[$display(frac(1, f ( 1 ))) =$]\_\_\_\_\_, which \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals minus 2")[$f^(− 1) ( 1 ) = − 2$], #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis equals 1")[$f ( − 2 ) = 1$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals minus 2")[$f^(− 1) ( 1 ) = − 2$], but #math.equation(block: false, alt: "the fraction 1 over f open parenthesis 1 close parenthesis equals 4")[$display(frac(1, f ( 1 ))) = 4$] ] + If #math.equation(block: false, alt: "z equals f open parenthesis w close parenthesis equals the fraction 1 over w plus 3")[$z = f ( w ) = display(frac(1, w + 3))$], find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$]. + Write two equations about the value of #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$], one using #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] and one using #math.equation(block: false, alt: "f")[$f$]. + Show that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$] is not equal to #math.equation(block: false, alt: "the fraction 1 over f open parenthesis 1 close parenthesis")[$display(frac(1, f ( 1 )))$]. #solutionbox[ + #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals minus 2")[$f^(− 1) ( 1 ) = − 2$], #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis equals 1")[$f ( − 2 ) = 1$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals minus 2")[$f^(− 1) ( 1 ) = − 2$], but #math.equation(block: false, alt: "the fraction 1 over f open parenthesis 1 close parenthesis equals 4")[$display(frac(1, f ( 1 ))) = 4$] ] We can use a graph of a function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] to find values of the inverse function #math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis")[$x = f^(− 1) ( y )$]. The figure below shows the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus 2")[$f ( x ) = x^(3) + 2$]. You already know how to evaluate a function from its graph: We start with the horizontal axis. For instance, to evaluate #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis")[$f ( − 2 )$], we find #math.equation(block: false, alt: "minus 2")[$− 2$] on the #math.equation(block: false, alt: "x")[$x$]-axis, move vertically to the point on the graph with #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$], in this case #math.equation(block: false, alt: "open parenthesis minus 2 , minus 6 close parenthesis")[$( − 2 , − 6 )$], and read the #math.equation(block: false, alt: "y")[$y$]-coordinate of the point. We see that #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis equals minus 6")[$f ( − 2 ) = − 6$]. To evaluate the inverse function, we start with the vertical axis. For example, to find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 10 close parenthesis")[$f^(− 1) ( 10 )$], we find #math.equation(block: false, alt: "10")[$10$] on the vertical axis and move horizontally to the point on the graph with #math.equation(block: false, alt: "y equals 10")[$y = 10$]. In this case, the point is #math.equation(block: false, alt: "open parenthesis 2 , 10 close parenthesis")[$( 2 , 10 )$], so #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 10 close parenthesis equals 2")[$f^(− 1) ( 10 ) = 2$]. #figure(figph[reading values of f-inverse from graph of f], alt: "reading values of f-inverse from graph of f", caption: none) #examplebox("Example 4")[][ The function #math.equation(block: false, alt: "C equals h open parenthesis F close parenthesis")[$C = h ( F )$] gives Celsius temperature as a function of Fahrenheit temperature. The graph of the function is shown below. Use the graph to evaluate #math.equation(block: false, alt: "h open parenthesis 68 close parenthesis")[$h ( 68 )$] and #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 10 close parenthesis")[$h^(− 1) ( 10 )$], and then explain their meaning in this context. #figure(figph[graph of Celsius vs Fahrenheit], alt: "graph of Celsius vs Fahrenheit", caption: none) #solutionbox[ To evaluate #math.equation(block: false, alt: "h open parenthesis 68 close parenthesis")[$h ( 68 )$], we find the input #math.equation(block: false, alt: "F equals 68")[$F = 68$] on the horizontal axis, then find the point on the graph with #math.equation(block: false, alt: "F equals 68")[$F = 68$] and read its vertical coordinate. We see that the point #math.equation(block: false, alt: "open parenthesis 68 , 20 close parenthesis")[$( 68 , 20 )$] lies on the graph, so #math.equation(block: false, alt: "h open parenthesis 68 close parenthesis equals 20")[$h ( 68 ) = 20$]. When the Fahrenheit temperature is #math.equation(block: false, alt: "68 degrees")[$68^(∘)$], the Celsius temperature is #math.equation(block: false, alt: "20 degrees")[$20^(∘)$]. The inverse function reverses the roles of input and output. Because #math.equation(block: false, alt: "C equals h open parenthesis F close parenthesis")[$C = h ( F )$], #math.equation(block: false, alt: "F equals h to the power minus 1 open parenthesis C close parenthesis")[$F = h^(− 1) ( C )$], so the inverse function gives us the Fahrenheit temperature if we know the Celsius temperature. In particular, #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 10 close parenthesis")[$h^(− 1) ( 10 )$] is the Fahrenheit temperature when the Celsius temperature is #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]. To use the graph of #math.equation(block: false, alt: "h")[$h$] to find values of #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$], we start with the vertical axis and find the point on the graph with #math.equation(block: false, alt: "C equals 10")[$C = 10$]. This point is #math.equation(block: false, alt: "open parenthesis 50 , 10 close parenthesis")[$( 50 , 10 )$], so #math.equation(block: false, alt: "F equals 50")[$F = 50$] when #math.equation(block: false, alt: "C equals 10")[$C = 10$], or #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 10 close parenthesis equals 50")[$h^(− 1) ( 10 ) = 50$]. When the Celsius temperature is #math.equation(block: false, alt: "10 degrees")[$10^(∘)$], the Fahrenheit temperature is #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]. ] ] Use the graph of #math.equation(block: false, alt: "h")[$h$] in Example. + Find #math.equation(block: false, alt: "h to the power minus 1 open parenthesis minus 10 close parenthesis")[$h^(− 1) ( − 10 )$]. #linebreak() #math.equation(block: false, alt: "h to the power minus 1 open parenthesis minus 10 close parenthesis equals")[$h^(− 1) ( − 10 ) =$]\_\_\_\_\_ + Does #math.equation(block: false, alt: "h to the power minus 1 open parenthesis minus 10 close parenthesis equals minus h to the power minus 1 open parenthesis 10 close parenthesis")[$h^(− 1) ( − 10 ) = − h^(− 1) ( 10 )$]? \_\_\_\_\_ + Select two equations, one using #math.equation(block: false, alt: "h")[$h$] and one using #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$], stating the Fahrenheit temperature when the Celsius temperature is #math.equation(block: false, alt: "0 degrees")[$0^(∘)$].- #math.equation(block: false, alt: "h open parenthesis 32 close parenthesis equals 0")[$h ( 32 ) = 0$] \_\_\_\_\_ - #math.equation(block: false, alt: "h open parenthesis 0 close parenthesis equals 32")[$h ( 0 ) = 32$] \_\_\_\_\_ - #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 32 close parenthesis equals 0")[$h^(− 1) ( 32 ) = 0$] \_\_\_\_\_ - #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 0 close parenthesis equals 32")[$h^(− 1) ( 0 ) = 32$] \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "14")[$14$]: On the graph of #math.equation(block: false, alt: "h")[$h$], when #math.equation(block: false, alt: "C equals minus 10")[$C = − 10$], #math.equation(block: false, alt: "F equals 14")[$F = 14$]. + No + #math.equation(block: false, alt: "h open parenthesis 32 close parenthesis equals 0")[$h ( 32 ) = 0$], #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 0 close parenthesis equals 32")[$h^(− 1) ( 0 ) = 32$] ] + Use the graph of #math.equation(block: false, alt: "h")[$h$] in Example to find #math.equation(block: false, alt: "h to the power minus 1 open parenthesis minus 10 close parenthesis")[$h^(− 1) ( − 10 )$]. + Does #math.equation(block: false, alt: "h to the power minus 1 open parenthesis minus 10 close parenthesis equals minus h to the power minus 1 open parenthesis 10 close parenthesis")[$h^(− 1) ( − 10 ) = − h^(− 1) ( 10 )$]? + Write two equations, one using #math.equation(block: false, alt: "h")[$h$] and one using #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$], stating the Fahrenheit temperature when the Celsius temperature is #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]. #solutionbox[ + #math.equation(block: false, alt: "14")[$14$]: On the graph of #math.equation(block: false, alt: "h")[$h$], when #math.equation(block: false, alt: "C equals minus 10")[$C = − 10$], #math.equation(block: false, alt: "F equals 14")[$F = 14$]. + No + #math.equation(block: false, alt: "h open parenthesis 32 close parenthesis equals 0")[$h ( 32 ) = 0$], #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 0 close parenthesis equals 32")[$h^(− 1) ( 0 ) = 32$] ] If the function #math.equation(block: false, alt: "f")[$f$] tells you how much money is in your account after #math.equation(block: false, alt: "t")[$t$] years, what does the function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] tell you? \_\_\_\_\_ If the function #math.equation(block: false, alt: "f")[$f$] tells you how much money is in your account after #math.equation(block: false, alt: "t")[$t$] years, what does the function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] tell you? === Graph of the Inverse Function In Example, we used a graph of #math.equation(block: false, alt: "h")[$h$] to read values of #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$]. But we can also plot the graph of #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$] itself. Because #math.equation(block: false, alt: "C")[$C$] is the input variable for #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$], we plot #math.equation(block: false, alt: "C")[$C$] on the horizontal axis and #math.equation(block: false, alt: "F")[$F$] on the vertical axis. To find some points on the graph of #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$], we interchange the coordinates of points on the graph of #math.equation(block: false, alt: "h")[$h$]. The graph of #math.equation(block: false, alt: "h to the power minus 1")[$h^(− 1)$] is shown at right. #figure(table( columns: 5, align: left, inset: 6pt, [#math.equation(block: false, alt: "C equals h open parenthesis F close parenthesis")[$C = h ( F )$]], [], [#math.equation(block: false, alt: "F equals h to the power minus 1 open parenthesis C close parenthesis")[$F = h^(− 1) ( C )$]], [], [], [#math.equation(block: false, alt: "F")[$F$]], [#math.equation(block: false, alt: "C")[$C$]], [$#hide($00000$)$], [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "F")[$F$]], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "minus 10")[$− 10$]], [], [#math.equation(block: false, alt: "minus 10")[$− 10$]], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "0")[$0$]], [], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "10")[$10$]], [], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "68")[$68$]], [#math.equation(block: false, alt: "20")[$20$]], [], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "68")[$68$]], )) #figure(figph[graph of Fahrenheit vs Celsius], alt: "graph of Fahrenheit vs Celsius", caption: none) #examplebox("Example 5")[][ The Park Service introduced a flock of #math.equation(block: false, alt: "12")[$12$] endangered pheasant into a wildlife preserve. After #math.equation(block: false, alt: "t")[$t$] years, the population of the flock was given by #math.equation(block: true, alt: "P equals f open parenthesis t close parenthesis equals 12 plus 2 t cubed")[$P = f ( t ) = 12 + 2 t^(3)$] + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 5 close bracket")[$[ 0 , 5 ]$]. + Find a formula for the inverse function, #math.equation(block: false, alt: "t equals f to the power minus 1 open parenthesis P close parenthesis")[$t = f^(− 1) ( P )$]. What is the meaning of the inverse function in this context? + Sketch a graph of the inverse function. #solutionbox[ + The graph of #math.equation(block: false, alt: "f")[$f$] is shown at right, with #math.equation(block: false, alt: "t")[$t$] on the horizontal axis and #math.equation(block: false, alt: "P")[$P$] on the vertical axis. #figure(figph[graph of pheasant population], alt: "graph of pheasant population", caption: none) + We solve #math.equation(block: false, alt: "P equals 12 plus 2 t cubed")[$" " P = 12 + 2 t^(3)$] for #math.equation(block: false, alt: "t")[$t$] in terms of #math.equation(block: false, alt: "P")[$P$]. #math.equation(block: true, alt: "2 t cubed, equals P minus 12, Substract 12 from both sides.; t cubed, equals the fraction P minus 12 over 2, Divide both sides by 2.; t, equals the cube root of the fraction P minus 12 over 2, Take cube roots.")[$2 t^(3) & = P − 12 & & "Substract 12 from both sides." \ t^(3) & = frac(P − 12, 2) & & "Divide both sides by 2." \ t & = root(3, frac(P − 12, 2)) & & "Take cube roots."$] The inverse function is #math.equation(block: false, alt: "t equals f to the power minus 1 open parenthesis P close parenthesis equals the cube root of the fraction P minus 12 over 2")[$" " t = f^(− 1) ( P ) = root(3, display(frac(P − 12, 2)))$]. It tells us the number of years it takes for the pheasant population to grow to size #math.equation(block: false, alt: "P")[$P$]. + The graph of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is shown below, with #math.equation(block: false, alt: "P")[$P$] on the horizontal axis and #math.equation(block: false, alt: "t")[$t$] on the vertical axis.#figure(figph[graph of time to achieve pheasant population], alt: "graph of time to achieve pheasant population", caption: none) ] ] The formula #math.equation(block: false, alt: "T equals f open parenthesis L close parenthesis equals 2 π the square root of the fraction L over 32")[$T = f ( L ) = 2 π sqrt(display(frac(L, 32)))$] gives the period in seconds, #math.equation(block: false, alt: "T")[$T$], of a pendulum as a function of its length in feet, #math.equation(block: false, alt: "L")[$L$]. + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 5 close bracket")[$[ 0 , 5 ]$]. + Find a formula for the inverse function, #math.equation(block: false, alt: "L equals f to the power minus 1 open parenthesis T close parenthesis")[$L = f^(− 1) ( T )$]. \[Enter "pi" to get #math.equation(block: false, alt: "π")[$π$].\] #linebreak() #math.equation(block: false, alt: "L equals f to the power minus 1 open parenthesis T close parenthesis equals")[$L = f^(− 1) ( T ) =$]\_\_\_\_\_ #linebreak() What is the meaning of the inverse function in this context? #linebreak() \_\_\_\_\_ + Sketch a graph of the inverse function. #solutionbox[ + A graph is below for both parts (a) and (c). + #math.equation(block: false, alt: "L equals f to the power minus 1 open parenthesis T close parenthesis equals the fraction 8 over π squared T squared .")[$L = f^(− 1) ( T ) = display(frac(8, π^(2))) T^(2) . " " " " " " " " " " " "$] #math.equation(block: false, alt: "f to the power minus 1")[$" " f^(− 1)$] gives the length of a pendulum as a function of its period. + See graph for part(a) below. ] #figure(figph[both inverse and reciprocal], alt: "both inverse and reciprocal", caption: none) The formula #math.equation(block: false, alt: "T equals f open parenthesis L close parenthesis equals 2 π the square root of the fraction L over 32")[$" " T = f ( L ) = 2 π sqrt(display(frac(L, 32)))$] gives the period in seconds, #math.equation(block: false, alt: "T")[$T$], of a pendulum as a function of its length in feet, #math.equation(block: false, alt: "L")[$L$]. + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 5 close bracket")[$[ 0 , 5 ]$]. + Find a formula for the inverse function, #math.equation(block: false, alt: "L equals f to the power minus 1 open parenthesis T close parenthesis")[$L = f^(− 1) ( T )$]. What is the meaning of the inverse function in this context? + Sketch a graph of the inverse function. #solutionbox[ + #figure(figph[both inverse and reciprocal], alt: "both inverse and reciprocal", caption: none) + #math.equation(block: false, alt: "L equals f to the power minus 1 open parenthesis T close parenthesis equals the fraction 8 over π squared T squared .")[$" " L = f^(− 1) ( T ) = display(frac(8, π^(2))) T^(2) . " " " " " " " " " " " "$] #math.equation(block: false, alt: "f to the power minus 1")[$" " f^(− 1)$] gives the length of a pendulum as a function of its period. + See graph for part(a). ] === When Is the Inverse a Function? We can always find the inverse of a function simply by interchanging the input and output variables. In the preceding examples, interchanging the variables created a new function. However, the inverse of a function does not always turn out to be a function itself. For example, to find the inverse of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals x squared")[$" " y = f ( x ) = x^(2)$], we solve for #math.equation(block: false, alt: "x")[$x$] to get #math.equation(block: false, alt: "x equals plus or minus the square root of y")[$" " x = ± sqrt(y)$]. When we regard #math.equation(block: false, alt: "y")[$y$] as the input and #math.equation(block: false, alt: "x")[$x$] as the output, the relationship does not describe a function. The graphs of #math.equation(block: false, alt: "f")[$f$] and its inverse are shown below. (Note that for the graph of the inverse, we plot #math.equation(block: false, alt: "y")[$y$] on the horizontal axis and #math.equation(block: false, alt: "x")[$x$] on the vertical axis.) Because the graph of the inverse does not pass the vertical line test, it is not a function. #figure(figph[graphs of x-squared and its inverse], alt: "graphs of x-squared and its inverse", caption: none) For many applications, it is important to know whether or not the inverse of #math.equation(block: false, alt: "f")[$f$] is a function. This can be determined from the graph of #math.equation(block: false, alt: "f")[$f$]. When we interchange the roles of the input and output variables, horizontal lines of the form #math.equation(block: false, alt: "y equals k")[$y = k$] become vertical lines. Thus, if the graph of the #emph[inverse] is going to pass the vertical line test, the graph of the #emph[original function] must pass the #strong[horizontal line test], namely, that no horizontal line should intersect the graph in more than one point. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Line Test] If no horizontal line intersects the graph of a function more than once, then its inverse is also a function. ] Notice that the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared")[$f ( x ) = x^(2)$] does not pass the horizontal line test, so we would not expect its inverse to be a function. #examplebox("Example 6")[][ Which of the functions shown below have inverses that are also functions? #figure(figph[graphs for vertical line test], alt: "graphs for vertical line test", caption: none) #solutionbox[ In each case, we apply the horizontal line test to determine whether the inverse is a function. Because no horizontal line intersects their graphs more than once, the functions pictured in figures (a) and (c) have inverses that are also functions. The functions in figures (b) and (d) do not have inverses that are functions. ] ] #figure(figph[graphs for vertical line test], alt: "graphs for vertical line test", caption: none) Which of the functions whose graphs are shown above have inverses that are also functions? + \_\_\_\_\_ + \_\_\_\_\_ + \_\_\_\_\_ + \_\_\_\_\_ #solutionbox[ (a) and (d) ] Which of the functions whose graphs are shown below have inverses that are also functions? #figure(figph[graphs for vertical line test], alt: "graphs for vertical line test", caption: none) #solutionbox[ (a) and (d) ] A function that passes the horizontal line test is called #strong[one-to-one], because each input has only one output and each output has only one input. A one-to-one function passes the horizontal line test as well as the vertical line test. With this terminology, we can state the following theorem. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[One-to-one Functions] The inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one. ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A function may have an inverse function even if we cannot find its formula. The function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power 5 plus x plus 1")[$" " f ( x ) = x^(5) + x + 1 " "$] shown in figure (a) is one-to-one, so it has an inverse function. We can even graph the inverse function, as shown in figure (b), by interchanging the coordinates of points on the graph of #math.equation(block: false, alt: "f")[$f$]. #figure(figph[graphs of f and f-inverse], alt: "graphs of f and f-inverse", caption: none) However, we cannot find a formula for the inverse function because we cannot solve the equation #math.equation(block: false, alt: "y equals x to the power 5 plus x plus 1")[$" " y = x^(5) + x + 1 " "$] for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "y")[$y$]. ] What does the term one-to-one mean? Give an example. \_\_\_\_\_ What does the term one-to-one mean? Give an example. === Mathematical Properties of the Inverse Function The inverse function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of the function #math.equation(block: false, alt: "f")[$f$]. In Example, the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 6 plus 2 t")[$" " f ( t ) = 6 + 2 t " "$] multiplies the input by #math.equation(block: false, alt: "2")[$2$] and then adds #math.equation(block: false, alt: "6")[$6$] to the result. The inverse function #math.equation(block: false, alt: "f to the power minus 1 open parenthesis H close parenthesis equals the fraction H minus 6 over 2")[$f^(− 1) ( H ) = display(frac(H − 6, 2))$] undoes those operations in reverse order: It subtracts #math.equation(block: false, alt: "6")[$6$] from the input and then divides the result by #math.equation(block: false, alt: "2")[$2$]. If we apply the function #math.equation(block: false, alt: "f")[$f$] to a given input value and then apply the function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] to the output from #math.equation(block: false, alt: "f")[$f$], the end result will be the original input value. For example, if we choose #math.equation(block: false, alt: "t equals 5")[$t = 5$] as an input value, we find that #math.equation(block: true, alt: "f open parenthesis 5 close parenthesis, equals 6 plus 2 open parenthesis 5 close parenthesis equals 16, Multiply by 2, then add 6.; and f to the power minus 1 open parenthesis 16 close parenthesis, equals the fraction 16 minus 6 over 2 equals 5, Subtract 6, then divide by 2.")[$f ( 5 ) & = 6 + 2 ( 5 ) = 16 & & " Multiply by 2, then add 6." \ "and " f^(− 1) ( 16 ) & = frac(16 − 6, 2) = 5 & & "Subtract 6, then divide by 2."$] #figure(figph[function and inverse diagram], alt: "function and inverse diagram", caption: none) We return to the original input value, #math.equation(block: false, alt: "5")[$5$], as illustrated above. An inverse function \_\_\_\_\_ #solutionbox[ interchanges the input and output. ] An inverse function + makes the output negative. + makes the input negative. + interchanges the input and output. + takes the reciprocal of the output. Example illustrates the fact that if #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], then #math.equation(block: false, alt: "f")[$f$] is also the inverse function for #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #examplebox("Example 7")[][ Consider the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus 2")[$" " f ( x ) = x^(3) + 2$] and its inverse, #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals the cube root of y minus 2")[$f^(− 1) ( y ) = root(3, y − 2)$]. + Show that the inverse function undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals 2")[$x = 2$]. + Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of the inverse function on #math.equation(block: false, alt: "y equals minus 25")[$y = − 25$]. #solutionbox[ + We first evaluate the function #math.equation(block: false, alt: "f")[$f$] for #math.equation(block: false, alt: "x equals 2")[$x = 2$]: #math.equation(block: true, alt: "f open parenthesis 2 close parenthesis equals 2 cubed plus 2 equals 10")[$f ( 2 ) = 2^(3) + 2 = 10$] Then we evaluate the inverse function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] at #math.equation(block: false, alt: "y equals 10")[$y = 10$]: #math.equation(block: true, alt: "f to the power minus 1 open parenthesis 10 close parenthesis equals the cube root of 10 minus 2 equals the cube root of 8 equals 2")[$f^(− 1) ( 10 ) = root(3, 10 − 2) = root(3, 8) = 2$] We started and ended with #math.equation(block: false, alt: "2")[$2$]. + We first evaluate the function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] for #math.equation(block: false, alt: "y equals minus 25")[$y = − 25$]: #math.equation(block: true, alt: "f to the power minus 1 open parenthesis minus 25 close parenthesis equals the cube root of minus 25 minus 2 equals minus 3")[$f^(− 1) ( − 25 ) = root(3, − 25 − 2) = − 3$] Then we evaluate the function #math.equation(block: false, alt: "f")[$f$] for #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$]: #math.equation(block: true, alt: "f open parenthesis minus 3 close parenthesis equals open parenthesis minus 3 close parenthesis cubed plus 2 equals minus 25")[$f ( − 3 ) = ( − 3 )^(3) + 2 = − 25$] We started and ended with #math.equation(block: false, alt: "minus 25")[$− 25$]. ] ] + Find a formula for the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x minus 1")[$f ( x ) = display(frac(2, x − 1))$] Use #math.equation(block: false, alt: "y")[$y$] as in input variable for #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals")[$f^(− 1) ( y ) =$]\_\_\_\_\_ + Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals x1")[$x = "x1"$]. #linebreak() #math.equation(block: false, alt: "f open parenthesis x1 close parenthesis equals")[$f ( "x1" ) =$]\_\_\_\_\_ and #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y1 close parenthesis equals")[$f^(− 1) ( "y1" ) =$]\_\_\_\_\_ + Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y2 close parenthesis equals")[$f^(− 1) ( "y2" ) =$]\_\_\_\_\_ and #math.equation(block: false, alt: "f open parenthesis x2 close parenthesis equals")[$f ( "x2" ) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals 1 plus the fraction 2 over y")[$f^(− 1) ( y ) = 1 + display(frac(2, y))$] + #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals 1")[$f ( 3 ) = 1$], and #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals 3")[$f^(− 1) ( 1 ) = 3$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis minus 2 close parenthesis equals 0")[$f^(− 1) ( − 2 ) = 0$] and #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals minus 2")[$f ( 0 ) = − 2$] ] + Find a formula for the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x minus 1")[$" " f ( x ) = display(frac(2, x − 1))$]. + Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals 3")[$x = 3$]. + Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$]. #solutionbox[ + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals 1 plus the fraction 2 over y")[$f^(− 1) ( y ) = 1 + display(frac(2, y))$] + #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals 1")[$f ( 3 ) = 1$], and #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals 3")[$f^(− 1) ( 1 ) = 3$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis minus 2 close parenthesis equals 0")[$f^(− 1) ( − 2 ) = 0$] and #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals minus 2")[$f ( 0 ) = − 2$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Functions and Inverse Functions] Suppose #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f")[$f$]. Then #math.equation(block: true, alt: "f to the power minus 1 open parenthesis f open parenthesis x close parenthesis close parenthesis equals x and f open parenthesis f to the power minus 1 open parenthesis y close parenthesis close parenthesis equals y")[$f^(− 1) ( f ( x ) ) = x " " " and " " " f ( f^(− 1) ( y ) ) = y$] as long as #math.equation(block: false, alt: "x")[$x$] is in the domain of #math.equation(block: false, alt: "f")[$f$], and #math.equation(block: false, alt: "y")[$y$] is in the domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. ] === Symmetry So far we have been careful to keep track of the input and output variables when we work with inverse functions. This is important when we are dealing with applications; the names of the variables are usually chosen because they have a meaning in the context of the application, and it would be confusing to change them. However, we can also study inverse functions purely as mathematical objects. There is a relationship between the graph of a function and the graph of its inverse that is easier to see if we plot them both on the same set of axes. A graph does not change if we change the names of the variables, so we can let #math.equation(block: false, alt: "x")[$x$] represent the input for both functions, and let #math.equation(block: false, alt: "y")[$y$] represent the output. Consider the function #math.equation(block: false, alt: "C equals h open parenthesis F close parenthesis")[$C = h ( F )$] from Example, and its inverse function, #math.equation(block: false, alt: "F equals h to the power minus 1 open parenthesis C close parenthesis")[$F = h^(− 1) ( C )$]. The formulas for these functions are #math.equation(block: true, alt: "C, equals h open parenthesis F close parenthesis equals the fraction 5 over 9 open parenthesis F minus 32 close parenthesis; F, equals h to the power minus 1 open parenthesis C close parenthesis equals 32 plus the fraction 9 over 5 C")[$C & = h ( F ) = frac(5, 9) ( F − 32 ) \ F & = h^(− 1) ( C ) = 32 + frac(9, 5) C$] But their graphs are the same if we write them as #math.equation(block: true, alt: "y, equals h open parenthesis x close parenthesis equals the fraction 5 over 9 open parenthesis x minus 32 close parenthesis; y, equals h to the power minus 1 open parenthesis x close parenthesis equals 32 plus the fraction 9 over 5 x")[$y & = h ( x ) = frac(5, 9) ( x − 32 ) \ y & = h^(− 1) ( x ) = 32 + frac(9, 5) x$] The graphs are shown below. #figure(figph[graphs of function and inverse], alt: "graphs of function and inverse", caption: none) Now, for every point #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] on the graph of #math.equation(block: false, alt: "f")[$f$], the point #math.equation(block: false, alt: "open parenthesis b , a close parenthesis")[$( b , a )$] is on the graph of the inverse function. Observe that the points #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] and #math.equation(block: false, alt: "open parenthesis b , a close parenthesis")[$( b , a )$] are always located symmetrically across the line #math.equation(block: false, alt: "y equals x")[$y = x$]. The graphs are #strong[symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$]], which means that if we were to place a mirror along the line #math.equation(block: false, alt: "y equals x")[$y = x$], each graph would be the reflection of the other. The symmetry this paragraph proves, kept live under a slider. The blue curve is f(x) = x³ + a; the red curve is its inverse, drawn by interchanging the coordinates of every point on f — which is all an inverse is. The two curves are mirror images across the dashed line y = x for every value of a. Watch what the slider does to each: raising a shifts f upward, and shifts the inverse by the same amount to the right — because swapping coordinates turns a vertical shift into a horizontal one, an input change into an output change.{"functions":\[{"color":"\#6a6a6a","expression":"x","expression2":"","id":"mirror","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#1f4e79","expression":"x^3 + a","expression2":"","id":"function-f","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","expression":"t^3 + a","expression2":"t","id":"inverse-f","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":2.5,"tMin":-2.5,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Shift a","max":2,"min":-2,"name":"a","step":0.1,"value":1}\],"title":"f(x) = x\\u00b3 + a and its inverse, mirrored in y = x","version":1,"viewport":{"centerX":0,"centerY":0,"scale":78}} The graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] are symmetric about \_\_\_\_\_ #solutionbox[ the line #math.equation(block: false, alt: "y equals x")[$y = x$]. ] The graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] are symmetric about + the #math.equation(block: false, alt: "x")[$x$]-axis + the #math.equation(block: false, alt: "y")[$y$]-axis + the asymptote. + the line #math.equation(block: false, alt: "y equals x")[$y = x$]. #examplebox("Example 8")[][ Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 the square root of x plus 4")[$f ( x ) = 2 sqrt(x + 4)$] on the domain #math.equation(block: false, alt: "open bracket minus 4 , 12 close bracket")[$[ − 4 , 12 ]$]. Graph its inverse function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on the same grid. #solutionbox[ The graph of #math.equation(block: false, alt: "f")[$f$] has the same shape as the graph of #math.equation(block: false, alt: "y equals the square root of x")[$y = sqrt(x)$], shifted #math.equation(block: false, alt: "4")[$4$] units to the left and stretched vertically by a factor of #math.equation(block: false, alt: "2")[$2$]. The figure at left below shows the graph of #math.equation(block: false, alt: "f")[$f$], along with a table of values. By interchanging the rows of the table, we obtain points on the graph of the inverse function, shown at right. #figure(figph[graphs and tables of a function and its inverse function], alt: "graphs and tables of a function and its inverse function", caption: none) If we use #math.equation(block: false, alt: "x")[$x$] as the input variable for both functions, and #math.equation(block: false, alt: "y")[$y$] as the output, we can graph #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on the same grid, as shown at right. The two graphs are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$]. #figure(figph[function and inverse on the same grid], alt: "function and inverse on the same grid", caption: none) ] ] Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus 2")[$" " f ( x ) = x^(3) + 2 " "$] and its inverse #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x minus 2")[$" " f^(− 1) ( x ) = root(3, x − 2) " "$] on the same set of axes, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$]. #solutionbox[ #figure(figph[cubic and inverse], alt: "cubic and inverse", caption: none) ] Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus 2")[$" " f ( x ) = x^(3) + 2 " "$] and its inverse #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x minus 2")[$" " f^(− 1) ( x ) = root(3, x − 2) " "$] on the same set of axes, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$]. #solutionbox[ #figure(figph[cubic and inverse], alt: "cubic and inverse", caption: none) ] Explain the difference between the meaning of the notation #math.equation(block: false, alt: "p to the power minus 1")[$p^(− 1)$] if #math.equation(block: false, alt: "p")[$p$] is a function or if #math.equation(block: false, alt: "p")[$p$] is a variable. \_\_\_\_\_ Explain the difference between the meaning of the notation #math.equation(block: false, alt: "p to the power minus 1")[$p^(− 1)$] if #math.equation(block: false, alt: "p")[$p$] is a function or if #math.equation(block: false, alt: "p")[$p$] is a variable. === Domain and Range When we interchange the input and output variables to obtain the inverse function, we interchange the domain and range of the function. For the functions graphed in Example, you can see that #math.equation(block: true, alt: "Domain open parenthesis f close parenthesis, equals open bracket minus 4 , 12 close bracket, and, Domain open parenthesis f to the power minus 1 close parenthesis, equals open bracket 0 , 8 close bracket; Range open parenthesis f close parenthesis, equals open bracket 0 , 8 close bracket, Range open parenthesis f minus 1 close parenthesis, equals open bracket minus 4 , 12 close bracket")[$"Domain" ( f ) & = [ − 4 , 12 ] & "and" & & "Domain" ( f^(− 1) ) & = [ 0 , 8 ] \ "Range" ( f ) & = [ 0 , 8 ] & & & "Range" ( f − 1 ) & = [ − 4 , 12 ]$] This relationship between the domain and range of a function and its inverse holds in general. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Domain and Range of the Inverse Function] If #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f ,")[$f ,$] then #math.equation(block: true, alt: "Domain open parenthesis f to the power minus 1 close parenthesis, equals Range open parenthesis f close parenthesis; Range open parenthesis f to the power minus 1 close parenthesis, equals Domain open parenthesis f close parenthesis")[$"Domain" ( f^(− 1) ) & = "Range" ( f ) \ "Range" ( f^(− 1) ) & = "Domain" ( f )$] ] #examplebox("Example 9")[][ + Graph the function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis equals the fraction 1 over x plus 3")[$y = f ( x ) = display(frac(1, x + 3))$] in the window #math.equation(block: true, alt: "Xmin, equals minus 6, Xmax equals 3.4; Ymin, equals minus 6, Ymax equals 3")[$"Xmin" & = − 6 & & "Xmax" = 3.4 \ "Ymin" & = − 6 & & "Ymax" = 3$] + Graph the inverse function in the same window, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$]. + State the domain and range of #math.equation(block: false, alt: "f")[$f$], and of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #solutionbox[ + The graph of #math.equation(block: false, alt: "f")[$f$] is shown below. It looks like the graph of #math.equation(block: false, alt: "y equals the fraction 1 over x")[$y = display(frac(1, x))$], shifted #math.equation(block: false, alt: "3")[$3$] units to the left. + To find the inverse function, we solve for #math.equation(block: false, alt: "x")[$x$]. Take the reciprocal of both sides of the equation. #math.equation(block: true, alt: "the fraction 1 over y, equals x plus 3, Subtract 3 from both sides.; x, equals the fraction 1 over y minus 3")[$display(frac(1, y)) & = x + 3 & & "Subtract 3 from both sides." \ x & = display(frac(1, y)) − 3$] The inverse function is #math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis equals the fraction 1 over y minus 3")[$" " x = f^(− 1) ( y ) = display(frac(1, y)) − 3$], or, using #math.equation(block: false, alt: "x")[$x$] for the input variable, #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x minus 3")[$" " f^(− 1) ( x ) = display(frac(1, x)) − 3$]. The graph of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] looks like the graph of #math.equation(block: false, alt: "y equals the fraction 1 over x")[$y = display(frac(1, x))$], shifted down #math.equation(block: false, alt: "3")[$3$] units, as shown below.#figure(figph[GC graph of function and inverse], alt: "GC graph of function and inverse", caption: none) + Because #math.equation(block: false, alt: "f")[$f$] is undefined at #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$], the domain of #math.equation(block: false, alt: "f")[$f$] is all real numbers except #math.equation(block: false, alt: "minus 3")[$− 3$]. The graph has a horizontal asymptote at #math.equation(block: false, alt: "y equals 0")[$y = 0$], so the range is all real numbers except #math.equation(block: false, alt: "0")[$0$]. #linebreak() The inverse function #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x minus 3")[$" " f^(− 1) ( x ) = display(frac(1, x)) − 3 " "$] is undefined at #math.equation(block: false, alt: "x equals 0")[$x = 0$], so its domain is all real numbers except #math.equation(block: false, alt: "0")[$0$]. The graph of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals minus 3")[$y = − 3$], so its range is all real numbers except #math.equation(block: false, alt: "minus 3")[$− 3$]. ] ] + Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x minus 1")[$f ( x ) = display(frac(2, x − 1))$] and its inverse function, #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] (which you found above in Practice 7, on the same set of axes, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$]. + State the domain and range of #math.equation(block: false, alt: "f")[$f$], and of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #linebreak() The domain of #math.equation(block: false, alt: "f")[$f$] includes all real numbers except \_\_\_\_\_, and the range of #math.equation(block: false, alt: "f")[$f$] includes all real number except \_\_\_\_\_ #linebreak() The domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] includes all real numbers except \_\_\_\_\_, and the range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] includes all real number except \_\_\_\_\_ #solutionbox[ + A graph is below. + Domain of #math.equation(block: false, alt: "f")[$f$]: all real numbers except #math.equation(block: false, alt: "1")[$1$], Range of #math.equation(block: false, alt: "f")[$f$]: all real numbers except #math.equation(block: false, alt: "0")[$0$], Domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers except #math.equation(block: false, alt: "0")[$0$], Range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers except #math.equation(block: false, alt: "1")[$1$] ] Another graph of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] #figure(figph[function and invese], alt: "function and invese", caption: none) + Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x minus 1")[$" " f ( x ) = display(frac(2, x − 1))$] and its inverse function, #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] (which you found above in Practice 7, on the same set of axes, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$]. + State the domain and range of #math.equation(block: false, alt: "f")[$f$], and of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. #solutionbox[ + #figure(figph[function and invese], alt: "function and invese", caption: none) + Domain of #math.equation(block: false, alt: "f")[$f$]: all real numbers except #math.equation(block: false, alt: "1")[$1$], Range of #math.equation(block: false, alt: "f")[$f$]: all real numbers except #math.equation(block: false, alt: "0")[$0$], Domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers except #math.equation(block: false, alt: "0")[$0$], Range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers except #math.equation(block: false, alt: "1")[$1$] ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Inverse function - Horizontal line test - One-to-one ==== CONCEPTS + The #strong[inverse] of a function describes the same relationship between two variables but interchanges the roles of the input and output. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Inverse Functions] If the inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function, then the inverse is denoted by the symbol #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$], and #math.equation(block: true, alt: "f to the power minus 1 open parenthesis b close parenthesis equals a if and only if f open parenthesis a close parenthesis equals b")[$f^(− 1) ( b ) = a " " " " " if and only if " " " " " f ( a ) = b$] ] + We can make a table of values for the inverse function, #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$], by interchanging the columns of a table for #math.equation(block: false, alt: "f")[$f$]. + If a function is defined by a formula in the form #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$], we can find a formula for its inverse function by solving the equation for #math.equation(block: false, alt: "x")[$x$] to get #math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis")[$x = f^(− 1) ( y )$]. + The inverse function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of the function #math.equation(block: false, alt: "f")[$f$], that is, if we apply the inverse function to the output of #math.equation(block: false, alt: "f")[$f$], we return to the original input value. + If #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], then #math.equation(block: false, alt: "f")[$f$] is also the inverse function for #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. + The graphs of #math.equation(block: false, alt: "f")[$f$] and its inverse function are #strong[symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$]] . + #strong[Horizontal line test]: If no horizontal line intersects the graph of a function more than once, then the inverse is also a function. + A function that passes the horizontal line test is called #strong[one-to-one]. + The inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Functions and Inverse Functions] Suppose #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f")[$f$]. Then #math.equation(block: true, alt: "f to the power minus 1 open parenthesis f open parenthesis x close parenthesis close parenthesis equals x and f open parenthesis f to the power minus 1 open parenthesis y close parenthesis close parenthesis equals y")[$f^(− 1) ( f ( x ) ) = x " " " and " " " f ( f^(− 1) ( y ) ) = y$]as long as #math.equation(block: false, alt: "x")[$x$] is in the domain of #math.equation(block: false, alt: "f")[$f$], and #math.equation(block: false, alt: "y")[$y$] is in the domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Domain and Range of the Inverse Function] If #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f ,")[$f ,$] then #math.equation(block: true, alt: "Domain open parenthesis f to the power minus 1 close parenthesis, equals Range open parenthesis f close parenthesis; Range open parenthesis f to the power minus 1 close parenthesis, equals Domain open parenthesis f close parenthesis")[$"Domain" ( f^(− 1) ) & = "Range" ( f ) \ "Range" ( f^(− 1) ) & = "Domain" ( f )$] ] ==== STUDY QUESTIONS + Explain how the terms #strong[inverse function], #strong[one-to-one], and #strong[horizontal line test] are related. + If you know that #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 3 close parenthesis equals minus 7")[$f^(− 1) ( 3 ) = − 7$], what can you say about the values of #math.equation(block: false, alt: "f")[$f$] ? + Explain how to use a graph of the function #math.equation(block: false, alt: "g")[$g$] to evaluate #math.equation(block: false, alt: "g to the power minus 1 open parenthesis 2 close parenthesis")[$g^(− 1) ( 2 )$]. + Evaluate #math.equation(block: false, alt: "f open parenthesis f to the power minus 1 open parenthesis 5 close parenthesis close parenthesis")[$f ( f^(− 1) ( 5 ) )$]. + Delbert says that if #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power 3 / 5")[$f ( x ) = x^(3 / 5)$], then #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals x to the power minus 3 / 5")[$f^(− 1) ( x ) = x^(− 3 / 5)$]. Is he correct? Why or why not? ==== SKILLS Practice each skill in the Homework problems listed. + Given certain function values, find values of the inverse function: \#1–4 + Interpret values of the inverse function: \#5–12 + Find a formula for the inverse function: \#9–22, 27–34 + Graph the inverse function: \#15 and 16, 23–34 + Find the domain and range of the inverse function: \#33 and 34 + Use the horizontal line test to identify one-to-one functions: \#35–42 === Homework 5.1 Let #math.equation(block: false, alt: "f open parenthesis minus 1 close parenthesis equals 0")[$f ( − 1 ) = 0$], #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals 1")[$f ( 0 ) = 1$], #math.equation(block: false, alt: "f open parenthesis 1 close parenthesis equals minus 2")[$f ( 1 ) = − 2$], and #math.equation(block: false, alt: "f open parenthesis 2 close parenthesis equals minus 1")[$f ( 2 ) = − 1$]. + Make a table of values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and another table for its inverse function. + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$] + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis minus 1 close parenthesis")[$f^(− 1) ( − 1 )$] + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]]), [#math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis")[$f^(− 1) ( y )$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis equals 0")[$f^(− 1) ( 1 ) = 0$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis minus 1 close parenthesis equals 2")[$f^(− 1) ( − 1 ) = 2$] Let #math.equation(block: false, alt: "f open parenthesis minus 2 close parenthesis equals 1")[$f ( − 2 ) = 1$], #math.equation(block: false, alt: "f open parenthesis minus 1 close parenthesis equals minus 2")[$f ( − 1 ) = − 2$], #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis equals 0")[$f ( 0 ) = 0$], and #math.equation(block: false, alt: "f open parenthesis 1 close parenthesis equals minus 1")[$f ( 1 ) = − 1$]. + Make a table of values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and another table for its inverse function. + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis minus 1 close parenthesis")[$f^(− 1) ( − 1 )$] + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed plus x plus 1")[$f ( x ) = x^(3) + x + 1$] + Make a table of values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and another table for its inverse function. + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 1 close parenthesis")[$f^(− 1) ( 1 )$] + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 3 close parenthesis")[$f^(− 1) ( 3 )$] + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "11")[$11$]], )) #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "11")[$11$]]), [#math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis")[$f^(− 1) ( y )$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) + #math.equation(block: false, alt: "f to the power 1 open parenthesis 1 close parenthesis equals 0")[$f^(1) ( 1 ) = 0$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 3 close parenthesis equals 1")[$f^(− 1) ( 3 ) = 1$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x to the power 5 plus x cubed plus 7")[$f ( x ) = x^(5) + x^(3) + 7$] + Make a table of values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and another table for its inverse function. + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 7 close parenthesis")[$f^(− 1) ( 7 )$] + Find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 5 close parenthesis")[$f^(− 1) ( 5 )$] For Problems 5-8, use the graph to evaluate each expression. An insurance investigator measures the length, #math.equation(block: false, alt: "d")[$d$], of the skid marks at an accident scene, in feet. The graph shows the function #math.equation(block: false, alt: "v equals f open parenthesis d close parenthesis")[$v = f ( d )$], which gives the velocity, #math.equation(block: false, alt: "v")[$v$] (mph), at which a car was traveling when it hit the brakes. #figure(figph[velocity vs skid marks length], alt: "velocity vs skid marks length", caption: none) + Use the graph to estimate #math.equation(block: false, alt: "f open parenthesis 60 close parenthesis")[$f ( 60 )$] and explain its meaning in this context. + Use the graph to estimate #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 60 close parenthesis")[$f^(− 1) ( 60 )$] and explain its meaning in this context. + #math.equation(block: false, alt: "f open parenthesis 60 close parenthesis approximately equals 38")[$f ( 60 ) ≈ 38$]. The car that left the #math.equation(block: false, alt: "60")[$60$]-foot skid marks was traveling at #math.equation(block: false, alt: "38")[$38$] mph. + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 60 close parenthesis approximately equals 150")[$f^(− 1) ( 60 ) ≈ 150$]. The car traveling at #math.equation(block: false, alt: "60")[$60$] mph left #math.equation(block: false, alt: "150")[$150$]-foot skid marks The weight, #math.equation(block: false, alt: "m")[$m$], of a missile launched from a catapult is a function of the distance, #math.equation(block: false, alt: "d")[$d$], to the target. The graph shows the function #math.equation(block: false, alt: "m equals f open parenthesis d close parenthesis")[$m = f ( d )$], where #math.equation(block: false, alt: "d")[$d$] is in meters and #math.equation(block: false, alt: "m")[$m$] is in kilograms. #figure(figph[mass vs target distance], alt: "mass vs target distance", caption: none) + Use the graph to estimate #math.equation(block: false, alt: "f open parenthesis 100 close parenthesis")[$f ( 100 )$] and explain its meaning in this context. + Use the graph to estimate #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 100 close parenthesis")[$f^(− 1) ( 100 )$] and explain its meaning in this context. After eating, the weight of a vampire bat drops steadily until its next meal. The graph shows the function #math.equation(block: false, alt: "W equals f open parenthesis t close parenthesis")[$W = f ( t )$], which gives the weight, #math.equation(block: false, alt: "W")[$W$], of the bat in grams #math.equation(block: false, alt: "t")[$t$] hours since its last meal. #figure(figph[bat weight vs time], alt: "bat weight vs time", caption: none) + Estimate the coordinates of the point of starvation. Include units in your answer. + Use the graph to estimate #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 90 close parenthesis")[$f^(− 1) ( 90 )$] and explain what it tells us about vampire bats. + #math.equation(block: false, alt: "open parenthesis 60 hours , 78 grams close parenthesis")[$( 60 " " "hours" , 78 " " "grams" )$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 90 close parenthesis approximately equals 19")[$f^(− 1) ( 90 ) ≈ 19$], so that the vampire bat's weight has dropped to #math.equation(block: false, alt: "90")[$90$] grams about #math.equation(block: false, alt: "19")[$19$] hours after its last meal. The amount of money, #math.equation(block: false, alt: "A")[$A$], in an interest-bearing savings account is a function of the number of years, #math.equation(block: false, alt: "t")[$t$], it remains in the account. The graph shows #math.equation(block: false, alt: "A equals f open parenthesis t close parenthesis")[$A = f ( t )$], where #math.equation(block: false, alt: "A")[$A$] is in thousands of dollars. #figure(figph[exponential growth], alt: "exponential growth", caption: none) + Use the graph to estimate #math.equation(block: false, alt: "f open parenthesis 30 close parenthesis")[$f ( 30 )$] and explain what it tells us about the account. + Use the graph to estimate #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 30 close parenthesis")[$f^(− 1) ( 30 )$] and explain what it tells us about the account. The function #math.equation(block: false, alt: "I equals g open parenthesis r close parenthesis equals open parenthesis 1 plus r close parenthesis to the power 5 minus 1")[$I = g ( r ) = ( 1 + r )^(5) − 1$] gives the interest, #math.equation(block: false, alt: "I")[$I$], that a dollar earns in #math.equation(block: false, alt: "5")[$5$] years in terms of the interest rate, #math.equation(block: false, alt: "r")[$r$]. + Evaluate #math.equation(block: false, alt: "g open parenthesis 0.05 close parenthesis")[$g ( 0.05 )$] and explain what it tells us about the interest. + Find the interest rate needed to earn #math.equation(block: false, alt: "$ 0.50")[$upright(\$) 0.50$] by substituting #math.equation(block: false, alt: "I equals 0.50")[$I = 0.50$] in the formula and solving for #math.equation(block: false, alt: "r")[$r$]. + Find a formula for the inverse function. + Write your answer to part (b) with inverse function notation. + #math.equation(block: false, alt: "g open parenthesis 0.05 close parenthesis equals 0.28")[$g ( 0.05 ) = 0.28$]. At #math.equation(block: false, alt: "5 %")[$5 upright(%)$] interest, #math.equation(block: false, alt: "$ 1")[$upright(\$) 1$] earns #math.equation(block: false, alt: "$ 0.28")[$upright(\$) 0.28$] interest in #math.equation(block: false, alt: "5")[$5$] years. + #math.equation(block: false, alt: "8.45 %")[$8.45 upright(%)$] + #math.equation(block: false, alt: "g to the power minus 1 open parenthesis I close parenthesis equals open parenthesis I plus 1 close parenthesis to the power 1 / 5 minus 1")[$g^(− 1) ( I ) = ( I + 1 )^(1 / 5) − 1$] + #math.equation(block: false, alt: "g to the power minus 1 open parenthesis 0.50 close parenthesis approximately equals 0.0845")[$g^(− 1) ( 0.50 ) ≈ 0.0845$] The function #math.equation(block: false, alt: "C equals h open parenthesis F close parenthesis equals the fraction 5 over 9 open parenthesis F minus 32 close parenthesis")[$C = h ( F ) = display(frac(5, 9)) ( F − 32 )$] gives the Celsius temperature #math.equation(block: false, alt: "C")[$C$] in terms of the Fahrenheit temperature #math.equation(block: false, alt: "F")[$F$]. + Evaluate #math.equation(block: false, alt: "h open parenthesis 104 close parenthesis")[$h ( 104 )$] and explain what it tells us about the temperature. + Find the Fahrenheit temperature of #math.equation(block: false, alt: "37 degrees")[$37^(∘)$] Celsius by substituting #math.equation(block: false, alt: "C equals 37")[$C = 37$] in the formula and solving for #math.equation(block: false, alt: "F")[$F$]. + Find a formula for the inverse function. + Write your answer to part (b) with inverse function notation. If you are flying in an airplane at an altitude of #math.equation(block: false, alt: "h")[$h$] miles, on a clear day you can see a distance of #math.equation(block: false, alt: "d")[$d$] miles to the horizon, where #math.equation(block: false, alt: "d equals f open parenthesis h close parenthesis equals the square root of 7920 h")[$d = f ( h ) = sqrt(7920 h)$]. + Evaluate #math.equation(block: false, alt: "f open parenthesis 0.5 close parenthesis")[$f ( 0.5 )$] and explain what it tells us about the horizon. + Find the altitude needed in order to see a distance of #math.equation(block: false, alt: "10")[$10$] mile by substituting #math.equation(block: false, alt: "d equals 10")[$d = 10$] in the formula and solving for #math.equation(block: false, alt: "h")[$h$]. + Find a formula for the inverse function. + Write your answer to part (b) with inverse function notation. + #math.equation(block: false, alt: "f open parenthesis 0.5 close parenthesis approximately equals 62.9")[$f ( 0.5 ) ≈ 62.9$]. At an altitude of #math.equation(block: false, alt: "0.5")[$0.5$] miles, you can see #math.equation(block: false, alt: "62.9")[$62.9$] miles to the horizon. + #math.equation(block: false, alt: "0.0126")[$0.0126$] mile, or #math.equation(block: false, alt: "66.7")[$66.7$] feet + #math.equation(block: false, alt: "h equals f to the power minus 1 open parenthesis d close parenthesis equals the fraction d squared over 7920")[$h = f^(− 1) ( d ) = display(frac(d^(2), 7920))$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 10 close parenthesis approximately equals 0.0126")[$f^(− 1) ( 10 ) ≈ 0.0126$] A moving ship creates waves that impede its own speed. The function #math.equation(block: false, alt: "v equals f open parenthesis L close parenthesis equals 1.3 the square root of L")[$v = f ( L ) = 1.3 sqrt(L)$] gives the ship's maximum speed in knots in terms of its length, #math.equation(block: false, alt: "L")[$L$], in feet. + Evaluate #math.equation(block: false, alt: "f open parenthesis 400 close parenthesis")[$f ( 400 )$] and explain what it tells us about the ship's speed. + Find the length needed for a maximum speed of #math.equation(block: false, alt: "35")[$35$] knots by substituting #math.equation(block: false, alt: "v equals 35")[$v = 35$] in the formula and solving for #math.equation(block: false, alt: "L")[$L$]. + Find a formula for the inverse function. + Write your answer to part (b) with inverse function notation. + Use the graph of #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the square root of 5 minus x")[$h ( x ) = sqrt(5 − x)$] to find #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 3 close parenthesis")[$h^(− 1) ( 3 )$]. + Find a formula for #math.equation(block: false, alt: "h to the power minus 1 open parenthesis x close parenthesis")[$h^(− 1) ( x )$] and evaluate #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 3 close parenthesis")[$h^(− 1) ( 3 )$]. #figure(figph[transformed square root], alt: "transformed square root", caption: none) + #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 3 close parenthesis approximately equals minus 4")[$h^(− 1) ( 3 ) ≈ − 4$] + #math.equation(block: false, alt: "h to the power minus 1 open parenthesis x close parenthesis equals 5 minus x squared")[$h^(− 1) ( x ) = 5 − x^(2)$]; #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 3 close parenthesis equals minus 4")[$" " h^(− 1) ( 3 ) = − 4$] + Use the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over 3 minus x")[$g ( x ) = display(frac(1, 3 − x))$] to find #math.equation(block: false, alt: "g to the power minus 1 open parenthesis minus 2 close parenthesis")[$g^(− 1) ( − 2 )$]. + Find a formula for #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis")[$g^(− 1) ( x )$] and evaluate #math.equation(block: false, alt: "g to the power minus 1 open parenthesis minus 2 close parenthesis")[$g^(− 1) ( − 2 )$]. #figure(figph[transformed reciprocal function], alt: "transformed reciprocal function", caption: none) + Find #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] for the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis x minus 2 close parenthesis cubed")[$f ( x ) = ( x − 2 )^(3)$]. + Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals 4")[$x = 4$]. + Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "x equals minus 8")[$x = − 8$]. + Graph the function and its inverse on the same grid, along with the graph of #math.equation(block: false, alt: "y equals x")[$y = x$]. + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals 3 the cube root of y plus 2")[$f^(− 1) ( y ) = 3 root(3, y) + 2$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis f open parenthesis 4 close parenthesis close parenthesis equals f to the power minus 1 open parenthesis 8 close parenthesis equals 4")[$f^(− 1) ( f ( 4 ) ) = f^(− 1) ( 8 ) = 4$] + #math.equation(block: false, alt: "f open parenthesis f to the power minus 1 open parenthesis minus 8 close parenthesis close parenthesis equals f open parenthesis 0 close parenthesis equals minus 8")[$f ( f^(− 1) ( − 8 ) ) = f ( 0 ) = − 8$] + #figure(figph[cubic and inverse], alt: "cubic and inverse", caption: none) + Find #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] for the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x plus 1")[$f ( x ) = display(frac(2, x + 1))$]. + Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals 3")[$x = 3$]. + Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. + Graph the function and its inverse on the same grid, along with the graph of #math.equation(block: false, alt: "y equals x")[$y = x$]. If #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals the fraction 2 over 3 t plus 1")[$F ( t ) = display(frac(2, 3)) t + 1$], find #math.equation(block: false, alt: "F to the power minus 1 open parenthesis 5 close parenthesis")[$F^(− 1) ( 5 )$]. #math.equation(block: true, alt: "6")[$6$] If #math.equation(block: false, alt: "G open parenthesis s close parenthesis equals the fraction s minus 3 over 4")[$G ( s ) = display(frac(s − 3, 4))$], find #math.equation(block: false, alt: "G to the power minus 1 open parenthesis minus 2 close parenthesis")[$G^(− 1) ( − 2 )$]. If #math.equation(block: false, alt: "m open parenthesis v close parenthesis equals 6 minus the fraction 2 over v")[$m ( v ) = 6 − display(frac(2, v))$], find #math.equation(block: false, alt: "m to the power minus 1 open parenthesis minus 3 close parenthesis")[$m^(− 1) ( − 3 )$]. #math.equation(block: true, alt: "the fraction 2 over 9")[$display(frac(2, 9))$] If #math.equation(block: false, alt: "p open parenthesis z close parenthesis equals 1 minus 2 z cubed")[$p ( z ) = 1 − 2 z^(3)$], find #math.equation(block: false, alt: "p to the power minus 1 open parenthesis 7 close parenthesis")[$p^(− 1) ( 7 )$]. If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x plus 2 over x minus 1")[$f ( x ) = display(frac(x + 2, x − 1))$], find #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 2 close parenthesis")[$f^(− 1) ( 2 )$]. #math.equation(block: true, alt: "4")[$4$] If #math.equation(block: false, alt: "g open parenthesis n close parenthesis equals the fraction 3 n plus 1 over n minus 3")[$g ( n ) = display(frac(3 n + 1, n − 3))$], find #math.equation(block: false, alt: "g to the power minus 1 open parenthesis minus 2 close parenthesis")[$g^(− 1) ( − 2 )$]. For Problems 23–26, + Use the graph to make a table of values for the function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. + Make a table of values and a graph of the inverse function. #figure(figph[line], alt: "line", caption: none) + #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "6")[$6$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "1200")[$1200$]], )) + #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "1200")[$1200$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "6")[$6$]], )) #figure(figph[line], alt: "line", caption: none) #figure(figph[line], alt: "line", caption: none) #figure(figph[curve], alt: "curve", caption: none) + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "100")[$100$]], )) + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) #figure(figph[line], alt: "line", caption: none) #figure(figph[root], alt: "root", caption: none) For Problems 27–32, + Find a formula for the inverse of the function. + Graph the function and its inverse on the same set of axes, along with the graph of #math.equation(block: false, alt: "y equals x")[$y = x$]. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 x minus 6")[$f ( x ) = 2 x − 6$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction x plus 6 over 2")[$f^(− 1) ( x ) = display(frac(x + 6, 2))$] + #figure(figph[function and inverse], alt: "function and inverse", caption: none) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 x minus 1")[$f ( x ) = 3 x − 1$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed plus 1")[$f ( x ) = x^(3) + 1$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x minus 1")[$f^(− 1) ( x ) = root(3, x − 1)$] + #figure(figph[function and inverse], alt: "function and inverse", caption: none) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the cube root of x plus 1")[$f ( x ) = root(3, x + 1)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 1")[$f ( x ) = display(frac(1, x − 1))$] + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x plus 1")[$f^(− 1) ( x ) = display(frac(1, x)) + 1$] + #figure(figph[function and inverse], alt: "function and inverse", caption: none) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 3")[$f ( x ) = display(frac(1, x)) − 3$] + Find the domain and range of the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the square root of 4 minus x")[$g ( x ) = sqrt(4 − x)$]. + Find a formula for #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis")[$g^(− 1) ( x )$]. + State the domain and range of #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis")[$g^(− 1) ( x )$]. + Graph #math.equation(block: false, alt: "g")[$g$] and #math.equation(block: false, alt: "g to the power minus 1")[$g^(− 1)$] on the same grid. + Domain: #math.equation(block: false, alt: "open parenthesis minus ∞ , 4 close bracket")[$( − upright(∞) , 4 ]$]; Range: #math.equation(block: false, alt: "open bracket 0 , ∞ close parenthesis")[$[ 0 , upright(∞) )$] + #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis equals 4 minus x squared")[$g^(− 1) ( x ) = 4 − x^(2)$] + Domain: #math.equation(block: false, alt: "open bracket 0 , ∞ close parenthesis")[$[ 0 , upright(∞) )$]; Range: #math.equation(block: false, alt: "open parenthesis minus ∞ , 4 close bracket")[$( − upright(∞) , 4 ]$] + #figure(figph[function and inverse], alt: "function and inverse", caption: none) + Find the domain and range of the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 8 minus the square root of x")[$g ( x ) = 8 − sqrt(x)$]. + Find a formula for #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis")[$g^(− 1) ( x )$]. + State the domain and range of #math.equation(block: false, alt: "g to the power minus 1 open parenthesis x close parenthesis")[$g^(− 1) ( x )$]. + Graph #math.equation(block: false, alt: "g")[$g$] and #math.equation(block: false, alt: "g to the power minus 1")[$g^(− 1)$] on the same grid. Which of the functions in Problems 35–42 have inverses that are also functions? #figure(figph[four curves], alt: "four curves", caption: none) (a) and (d) #figure(figph[four curves], alt: "four curves", caption: none) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x")[$f ( x ) = x$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared")[$f ( x ) = x^(2)$] (a) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$f ( x ) = x^(3)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals vertical bar x vertical bar")[$f ( x ) = | x |$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x")[$f ( x ) = display(frac(1, x))$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x squared")[$f ( x ) = display(frac(1, x^(2)))$] (a) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the square root of x")[$f ( x ) = sqrt(x)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the cube root of x")[$f ( x ) = root(3, x)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( display(frac(1, 2)) ), t: x)$] (a) and (b) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus x squared")[$f ( x ) = x^(3) + x^(2)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed plus x")[$f ( x ) = x^(3) + x$] Find a formula for each function shown in (a)–(d). Then match each function with its inverse from I–IV. #figure(figph[two columns of four graphs], alt: "two columns of four graphs", caption: none) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 plus 2 x")[$f ( x ) = 4 + 2 x$]; IV + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 minus the fraction x over 2")[$f ( x ) = 2 − display(frac(x, 2))$]; III + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 4 minus 2 x")[$f ( x ) = − 4 − 2 x$]; I + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x over 2")[$f ( x ) = display(frac(x, 2))$]; II Find a formula for each function shown in (a)–(d). Then match each function with its inverse from I–IV. #figure(figph[two columns of four graphs], alt: "two columns of four graphs", caption: none) For Problems 45 and 46, use the graph of #math.equation(block: false, alt: "f")[$f$] to match the other graphs with the appropriate function. (#emph[Hint]: Look at the coordinates of some specific points.) #figure(figph[increasing curve], alt: "increasing curve", caption: none) + #math.equation(block: false, alt: "minus f")[$− f$] + #math.equation(block: false, alt: "the fraction 1 over f")[$display(frac(1, f))$] + #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + III + II + I #figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "minus f")[$− f$] + #math.equation(block: false, alt: "the fraction 1 over f")[$display(frac(1, f))$] + #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none)