#set document(title: "10.1 Finding Composite and Inverse Functions", author: "OpenStax / XYZ Homework")
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== 10.1#h(0.6em)Finding Composite and Inverse Functions

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Before you get started, take this readiness quiz.

If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x minus 3")[$f ( x ) = 2 x − 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared plus 2 x minus 3 ,")[$g ( x ) = x^(2) + 2 x − 3 ,$] find #math.equation(block: false, alt: "f open parenthesis 4 close parenthesis .")[$f ( 4 ) .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
ⓐ #math.equation(block: false, alt: "f 4 equals 5")[$f 4 = 5$]; ⓑ #math.equation(block: false, alt: "g f 4 equals 32")[$g f 4 = 32$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve for #math.equation(block: false, alt: "x ,")[$x ,$] #math.equation(block: false, alt: "3 x plus 2 y equals 12 .")[$3 x + 2 y = 12 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x equals minus the fraction 2 over 3 y plus 4")[$x = − frac(2, 3) y + 4$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Simplify: #math.equation(block: false, alt: "5 the fraction open parenthesis x plus 4 close parenthesis over 5 minus 4 .")[$5 frac(( x + 4 ), 5) − 4 .$] #linebreak() 
If you missed this problem, review .

#solutionbox[
#math.equation(block: true, alt: "x")[$x$]
]
]

In this chapter, we will introduce two new types of functions, exponential functions and logarithmic functions. These functions are used extensively in business and the sciences as we will see.


=== Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called #strong[composition]. In composition, the output of one function is the input of a second function. For functions #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g ,")[$g ,$] the composition is written #math.equation(block: false, alt: "f ∘ g")[$f ∘ g$] and is defined by #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis equals f open parenthesis g open parenthesis x close parenthesis close parenthesis .")[$( f ∘ g ) ( x ) = f ( g ( x ) ) .$]

We read #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$f ( g ( x ) )$] as #math.equation(block: false, alt: "“ f")[$"“" f$] of #math.equation(block: false, alt: "g")[$g$] of #math.equation(block: false, alt: "x .”")[$x ".”"$]

#figure(figph[This figure shows x as the input to a box denoted as function g with g of x as the output of the box. Then, g of x is the input to a box denoted as function f with f of g of x as the output of the box.], alt: "This figure shows x as the input to a box denoted as function g with g of x as the output of the box. Then, g of x is the input to a box denoted as function f with f of g of x as the output of the box.", caption: none)

To do a composition, the output of the first function, #math.equation(block: false, alt: "g open parenthesis x close parenthesis ,")[$g ( x ) ,$] becomes the input of the second function, #emph[f], and so we must be sure that it is part of the domain of #emph[f].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Composition of Functions]

The composition of functions #emph[f] and #emph[g] is written #math.equation(block: false, alt: "f ∘ g")[$f ∘ g$] and is defined by

#math.equation(block: true, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis equals f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$( f ∘ g ) ( x ) = f ( g ( x ) )$]We read #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$f ( g ( x ) )$] as #math.equation(block: false, alt: "f")[$f$] of #math.equation(block: false, alt: "g")[$g$] of #emph[x].
]

We have actually used composition without using the notation many times before. When we graphed quadratic functions using translations, we were composing functions. For example, if we first graphed #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared")[$g ( x ) = x^(2)$] as a parabola and then shifted it down vertically four units, we were using the composition defined by #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis equals f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$( f ∘ g ) ( x ) = f ( g ( x ) )$] where #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x minus 4 .")[$f ( x ) = x − 4 .$]

#figure(figph[This figure shows x as the input to a box denoted as g of x equals x squared with x squared as the output of the box. Then, x squared is the input to a box denoted as f of x equals x minus 4 with f of g of x equals x squared minus 4 as the output of the box.], alt: "This figure shows x as the input to a box denoted as g of x equals x squared with x squared as the output of the box. Then, x squared is the input to a box denoted as f of x equals x minus 4 with f of g of x equals x squared minus 4 as the output of the box.", caption: none)

The next example will demonstrate that #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis ,")[$( f ∘ g ) ( x ) ,$] #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis x close parenthesis")[$( g ∘ f ) ( x )$] and #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis")[$( f · g ) ( x )$] usually result in different outputs.

#examplebox("Example 1")[][
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x minus 5")[$f ( x ) = 4 x − 5$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x plus 3 ,")[$g ( x ) = 2 x + 3 ,$] find: ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis ,")[$( f ∘ g ) ( x ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis x close parenthesis ,")[$( g ∘ f ) ( x ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis .")[$( f · g ) ( x ) .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the definition of #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis .")[$( f ∘ g ) ( x ) .$]], [#figure(figph[The image shows the definition of a composite function, stating that (f o g)(x) is equal to f(g(x)). This mathematical notation explains how to apply one function to the result of another function.], alt: "The image shows the definition of a composite function, stating that (f o g)(x) is equal to f(g(x)). This mathematical notation explains how to apply one function to the result of another function.", caption: none)]),
  [#figure(figph[The text 'Substitute 2x + 3 for g(x)' is displayed in black font, with '2x + 3' highlighted in red font, against a white background.], alt: "The text 'Substitute 2x + 3 for g(x)' is displayed in black font, with '2x + 3' highlighted in red font, against a white background.", caption: none)], [#figure(figph[Mathematical notation demonstrating the composition of two functions, (f o g)(x), which is further expressed as f(2x + 3), implying that the inner function g(x) is equal to 2x + 3.], alt: "Mathematical notation demonstrating the composition of two functions, (f o g)(x), which is further expressed as f(2x + 3), implying that the inner function g(x) is equal to 2x + 3.", caption: none)],
  [#figure(figph[A mathematical problem asking to find f(2x + 3) given that f(x) = 4x - 5. The expression '2x + 3' is highlighted in red, as is the 'x' in 'f(x)' and '4x' to denote variable substitution.], alt: "A mathematical problem asking to find f(2x + 3) given that f(x) = 4x - 5. The expression '2x + 3' is highlighted in red, as is the 'x' in 'f(x)' and '4x' to denote variable substitution.", caption: none)], [#figure(figph[The image shows a mathematical equation for the composition of two functions, f and g, denoted as (f o g)(x) = 4(2x + 3) - 5. The expression 2x+3 is highlighted in red.], alt: "The image shows a mathematical equation for the composition of two functions, f and g, denoted as (f o g)(x) = 4(2x + 3) - 5. The expression 2x+3 is highlighted in red.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation shows the composition of two functions, (f ∘ g)(x), which equals the expression 8x + 12 - 5. This simplifies to (f ∘ g)(x) = 8x + 7.], alt: "A mathematical equation shows the composition of two functions, (f ∘ g)(x), which equals the expression 8x + 12 - 5. This simplifies to (f ∘ g)(x) = 8x + 7.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the mathematical expression for a composite function, stating that (f o g)(x) = 8x + 7, where 'f' and 'g' are functions and 'o' denotes their composition.], alt: "The image shows the mathematical expression for a composite function, stating that (f o g)(x) = 8x + 7, where 'f' and 'g' are functions and 'o' denotes their composition.", caption: none)],
))

#linebreak() 
ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the definition of #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis .")[$( f ∘ g ) ( x ) .$]], [#figure(figph[The definition of a composite function: (g o f)(x) = g(f(x)).], alt: "The definition of a composite function: (g o f)(x) = g(f(x)).", caption: none)]),
  [#figure(figph[Substitute 4x - 5 for f(x).], alt: "Substitute 4x - 5 for f(x).", caption: none)], [#figure(figph[The image displays the composition of functions, showing (g o f)(x) as g(4x - 5).], alt: "The image displays the composition of functions, showing (g o f)(x) as g(4x - 5).", caption: none)],
  [#figure(figph[Find g(4x - 5) where g(x) = 2x + 3.], alt: "Find g(4x - 5) where g(x) = 2x + 3.", caption: none)], [#figure(figph[A mathematical equation showing the composition of functions (g o f)(x) = 2(4x - 5) + 3, with the term 4x - 5 highlighted in red.], alt: "A mathematical equation showing the composition of functions (g o f)(x) = 2(4x - 5) + 3, with the term 4x - 5 highlighted in red.", caption: none)],
  [Distribute.], [#figure(figph[A mathematical equation shows a composite function (g o f)(x) equals 8x - 10 + 3.], alt: "A mathematical equation shows a composite function (g o f)(x) equals 8x - 10 + 3.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the composition of two functions (g o f)(x) equal to 8x - 7.], alt: "The image shows the composition of two functions (g o f)(x) equal to 8x - 7.", caption: none)],
))

Notice the difference in the result in part ⓐ and part ⓑ.

ⓒ Notice that #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis")[$( f · g ) ( x )$] is different than #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis .")[$( f ∘ g ) ( x ) .$] In part ⓐ we did the composition of the functions. Now in part ⓒ we are not composing them, we are multiplying them. #linebreak()  #linebreak() 
#math.equation(block: false, alt: "Use the definition of open parenthesis f times g close parenthesis open parenthesis x close parenthesis ., open parenthesis f times g close parenthesis open parenthesis x close parenthesis equals f open parenthesis x close parenthesis times g open parenthesis x close parenthesis; Substitute f open parenthesis x close parenthesis equals 4 x minus 5 and g open parenthesis x close parenthesis equals 2 x plus 3 ., open parenthesis f times g close parenthesis open parenthesis x close parenthesis equals open parenthesis 4 x minus 5 close parenthesis times open parenthesis 2 x plus 3 close parenthesis; Multiply., open parenthesis f times g close parenthesis open parenthesis x close parenthesis equals 8 x squared plus 2 x minus 15")[$"Use the definition of" #h(0.2em) ( f · g ) ( x ) . &  &  &  &  & ( f · g ) ( x ) = f ( x ) · g ( x ) \ "Substitute" #h(0.2em) f ( x ) = 4 x − 5 #h(0.2em) "and" #h(0.2em) g ( x ) = 2 x + 3 . &  &  &  &  & ( f · g ) ( x ) = ( 4 x − 5 ) · ( 2 x + 3 ) \ "Multiply." &  &  &  &  & ( f · g ) ( x ) = 8 x^(2) + 2 x − 15$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x minus 2")[$f ( x ) = 3 x − 2$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 x plus 1 ,")[$g ( x ) = 5 x + 1 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis")[$( f ∘ g ) ( x )$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis x close parenthesis")[$( g ∘ f ) ( x )$] ⓒ #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis")[$( f · g ) ( x )$].

#solutionbox[
ⓐ #math.equation(block: false, alt: "15 x plus 1")[$15 x + 1$] ⓑ #math.equation(block: false, alt: "15 x minus 9")[$15 x − 9$] #linebreak() ⓒ #math.equation(block: false, alt: "15 x squared minus 7 x minus 2")[$15 x^(2) − 7 x − 2$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x minus 3 ,")[$f ( x ) = 4 x − 3 ,$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 6 x minus 5 ,")[$g ( x ) = 6 x − 5 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis ,")[$( f ∘ g ) ( x ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis x close parenthesis ,")[$( g ∘ f ) ( x ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis .")[$( f · g ) ( x ) .$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "24 x minus 23")[$24 x − 23$] ⓑ #math.equation(block: false, alt: "24 x minus 23")[$24 x − 23$] #linebreak() ⓒ #math.equation(block: false, alt: "24 x squared minus 38 x plus 15")[$24 x^(2) − 38 x + 15$]
]
]

In the next example we will evaluate a composition for a specific value.

#examplebox("Example 2")[][
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 4 ,")[$f ( x ) = x^(2) − 4 ,$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x plus 2 ,")[$g ( x ) = 3 x + 2 ,$] find: ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −3 close parenthesis ,")[$( f ∘ g ) ( −3 ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −1 close parenthesis ,")[$( g ∘ f ) ( −1 ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis 2 close parenthesis .")[$( f ∘ f ) ( 2 ) .$]

#solutionbox[
ⓐ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the definition of #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −3 close parenthesis .")[$( f ∘ g ) ( −3 ) .$]], [#figure(figph[A mathematical expression showing the composition of two functions, f and g, evaluated at -3. The notation (f o g)(-3) is explicitly shown to be equal to f(g(-3)).], alt: "A mathematical expression showing the composition of two functions, f and g, evaluated at -3. The notation (f o g)(-3) is explicitly shown to be equal to f(g(-3)).", caption: none)]),
  [#figure(figph[The image displays a mathematical problem asking to find the value of the function g at x = -3, given that g(x) = 3x + 2.], alt: "The image displays a mathematical problem asking to find the value of the function g at x = -3, given that g(x) = 3x + 2.", caption: none)], [#figure(figph[A mathematical equation shows the composition of functions (f o g)(-3) expanded as f(3 \* (-3) + 2). The number -3 is highlighted in red within the expanded expression, indicating its role in the calculation.], alt: "A mathematical equation shows the composition of functions (f o g)(-3) expanded as f(3 * (-3) + 2). The number -3 is highlighted in red within the expanded expression, indicating its role in the calculation.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation showing the composition of functions, stating that (f composed with g) evaluated at -3 is equal to f evaluated at -7, written as (f ∘ g)(-3) = f(-7).], alt: "A mathematical equation showing the composition of functions, stating that (f composed with g) evaluated at -3 is equal to f evaluated at -7, written as (f ∘ g)(-3) = f(-7).", caption: none)],
  [#figure(figph[The image shows a mathematical problem that asks to 'Find f(-7) where f(x) = x^2 - 4.'], alt: "The image shows a mathematical problem that asks to 'Find f(-7) where f(x) = x^2 - 4.'", caption: none)], [#figure(figph[The image shows the mathematical expression for a composite function: (f o g)(-3) = (-7)^2 - 4. The value -7 is highlighted in red, indicating a substitution or an intermediate step in the calculation.], alt: "The image shows the mathematical expression for a composite function: (f o g)(-3) = (-7)^2 - 4. The value -7 is highlighted in red, indicating a substitution or an intermediate step in the calculation.", caption: none)],
  [Simplify.], [#figure(figph[The image shows the mathematical expression for a composite function: (f o g)(-3) = 45.], alt: "The image shows the mathematical expression for a composite function: (f o g)(-3) = 45.", caption: none)],
))

#linebreak() 
ⓑ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the definition of #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −1 close parenthesis .")[$( g ∘ f ) ( −1 ) .$]], [#figure(figph[Composite function notation showing that (g composed with f) of negative 1 equals g of f of negative 1.], alt: "Composite function notation showing that (g composed with f) of negative 1 equals g of f of negative 1.", caption: none)]),
  [#figure(figph[A mathematics problem asks to find the value of f(-1) given the function f(x) = x^2 - 4. The text is clear and readable on a white background.], alt: "A mathematics problem asks to find the value of f(-1) given the function f(x) = x^2 - 4. The text is clear and readable on a white background.", caption: none)], [#figure(figph[A mathematical equation demonstrating function composition, (g composed with f) at -1, which simplifies to g evaluated at the expression (-1)^2 - 4. The -1 in the exponent is highlighted.], alt: "A mathematical equation demonstrating function composition, (g composed with f) at -1, which simplifies to g evaluated at the expression (-1)^2 - 4. The -1 in the exponent is highlighted.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation showing the composition of two functions, (g o f)(-1) = g(-3).], alt: "A mathematical equation showing the composition of two functions, (g o f)(-1) = g(-3).", caption: none)],
  [#figure(figph[A math problem states: 'Find g(-3) where g(x) = 3x + 2.' The number -3 and the variable x in g(x) and 3x are highlighted in red.], alt: "A math problem states: 'Find g(-3) where g(x) = 3x + 2.' The number -3 and the variable x in g(x) and 3x are highlighted in red.", caption: none)], [#figure(figph[The image shows the mathematical equation for a composite function evaluation, specifically (g  f)(-1) = 3(-3) + 2, where -3 is highlighted in red.], alt: "The image shows the mathematical equation for a composite function evaluation, specifically (g f)(-1) = 3(-3) + 2, where -3 is highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression showing the composition of two functions: (g o f)(-1) = -7. The expression is rendered in black text on a white background, representing a function evaluation resulting in a negative integer.], alt: "A mathematical expression showing the composition of two functions: (g o f)(-1) = -7. The expression is rendered in black text on a white background, representing a function evaluation resulting in a negative integer.", caption: none)],
))

#linebreak() 
ⓒ  #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Use the definition of #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis 2 close parenthesis .")[$( f ∘ f ) ( 2 ) .$]], [#figure(figph[(f o f)(2) = f(f(2)) is shown, illustrating the composition of function f with itself, evaluated at x=2.], alt: "(f o f)(2) = f(f(2)) is shown, illustrating the composition of function f with itself, evaluated at x=2.", caption: none)]),
  [#figure(figph[A mathematical problem asking to find the value of f(2) given the function f(x) = x^2 - 4. The number 2 and the variable x in f(x) are highlighted in red.], alt: "A mathematical problem asking to find the value of f(2) given the function f(x) = x^2 - 4. The number 2 and the variable x in f(x) are highlighted in red.", caption: none)], [#figure(figph[A mathematical equation shows the composite function (f o f)(2) is equal to f(2^2 - 4). The number 2 in the exponent of 2^2 is highlighted in red, indicating a specific part of the expression.], alt: "A mathematical equation shows the composite function (f o f)(2) is equal to f(2^2 - 4). The number 2 in the exponent of 2^2 is highlighted in red, indicating a specific part of the expression.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows the composition of a function with itself evaluated at 2, equaling the function evaluated at 0: (f o f)(2) = f(0).], alt: "A mathematical equation shows the composition of a function with itself evaluated at 2, equaling the function evaluated at 0: (f o f)(2) = f(0).", caption: none)],
  [#figure(figph[The image shows the mathematical problem: Find f(0) where f(x) = x^2 - 4.], alt: "The image shows the mathematical problem: Find f(0) where f(x) = x^2 - 4.", caption: none)], [#figure(figph[A mathematical equation showing the composition of a function with itself evaluated at 2, setting (f o f)(2) equal to 0 squared minus 4. The number 0 is highlighted in red.], alt: "A mathematical equation showing the composition of a function with itself evaluated at 2, setting (f o f)(2) equal to 0 squared minus 4. The number 0 is highlighted in red.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical expression reads (f o f)(2) = -4, representing the composite function f of f evaluated at 2 equals -4. The notation is clear and centrally positioned.], alt: "A mathematical expression reads (f o f)(2) = -4, representing the composite function f of f evaluated at 2 equals -4. The notation is clear and centrally positioned.", caption: none)],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 9 ,")[$f ( x ) = x^(2) − 9 ,$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x plus 5 ,")[$g ( x ) = 2 x + 5 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −2 close parenthesis ,")[$( f ∘ g ) ( −2 ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −3 close parenthesis ,")[$( g ∘ f ) ( −3 ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis 4 close parenthesis .")[$( f ∘ f ) ( 4 ) .$]

#solutionbox[
ⓐ –8 ⓑ 5 ⓒ 40
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 1 ,")[$f ( x ) = x^(2) + 1 ,$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x minus 5 ,")[$g ( x ) = 3 x − 5 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −1 close parenthesis ,")[$( f ∘ g ) ( −1 ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis 2 close parenthesis ,")[$( g ∘ f ) ( 2 ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis −1 close parenthesis .")[$( f ∘ f ) ( −1 ) .$]

#solutionbox[
ⓐ 65 ⓑ 10 ⓒ 5
]
]


=== Determine Whether a Function is One-to-One

When we first introduced functions, we said a #strong[function] is a relation that assigns to each element in its domain exactly one element in the range. For each ordered pair in the relation, each #emph[x]-value is matched with only one #emph[y]-value.

We used the birthday example to help us understand the definition. Every person has a birthday, but no one has two birthdays and it is okay for two people to share a birthday. Since each person has exactly one birthday, that relation is a function.

#figure(figph[This figure shows two tables. To the left is the table labeled Name, which from top to bottom reads Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, and Danny. The table on the right is labeled Birthday, which from top to bottom reads January 12, February 3, April 25, May 10, May 23, July 24, August 2, and September 15. There are arrows going from Alison to April 25, Penelope to May 23, June to August 2, Gregory to September 15, Geoffrey to January 12, Lauren to May 10, Stephen to July 24, Alice to February 3, Liz to July 24, and Danny to no birthday.], alt: "This figure shows two tables. To the left is the table labeled Name, which from top to bottom reads Alison, Penelope, June, Gregory, Geoffrey, Lauren, Stephen, Alice, Liz, and Danny. The table on the right is labeled Birthday, which from top to bottom reads January 12, February 3, April 25, May 10, May 23, July 24, August 2, and September 15. There are arrows going from Alison to April 25, Penelope to May 23, June to August 2, Gregory to September 15, Geoffrey to January 12, Lauren to May 10, Stephen to July 24, Alice to February 3, Liz to July 24, and Danny to no birthday.", caption: none)

A function is #strong[one-to-one] if each value in the range has exactly one element in the domain. For each ordered pair in the function, each #emph[y]-value is matched with only one #emph[x]-value.

Our example of the birthday relation is not a one-to-one function. Two people can share the same birthday. The range value August 2 is the birthday of Liz and June, and so one range value has two domain values. Therefore, the function is not one-to-one.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[One-to-One Function]

A function is #strong[one-to-one] if each value in the range corresponds to one element in the domain. For each ordered pair in the function, each #emph[y]-value is matched with only one #emph[x]-value. There are no repeated #emph[y]-values.
]

#examplebox("Example 3")[][
For each set of ordered pairs, determine if it represents a function and, if so, if the function is one-to-one.

ⓐ #math.equation(block: false, alt: "{ open parenthesis −3 , 27 close parenthesis , open parenthesis −2 , 8 close parenthesis , open parenthesis −1 , 1 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 2 , 8 close parenthesis , open parenthesis 3 , 27 close parenthesis }")[$\{ ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) \}$] and ⓑ #math.equation(block: false, alt: "{ open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 9 , 3 close parenthesis , open parenthesis 16 , 4 close parenthesis } .")[$\{ ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) , ( 16 , 4 ) \} .$]

#solutionbox[
ⓐ  #linebreak() 
#math.equation(block: false, alt: "{ open parenthesis −3 , 27 close parenthesis , open parenthesis −2 , 8 close parenthesis , open parenthesis −1 , 1 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 2 , 8 close parenthesis , open parenthesis 3 , 27 close parenthesis }")[$&  &  &  & #h(5em) \{ ( −3 , 27 ) , ( −2 , 8 ) , ( −1 , 1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 8 ) , ( 3 , 27 ) \}$]

Each #emph[x]-value is matched with only one #emph[y]-value. So this relation is a function.

But each #emph[y]-value is not paired with only one #emph[x]-value, #math.equation(block: false, alt: "open parenthesis −3 , 27 close parenthesis")[$( −3 , 27 )$] and #math.equation(block: false, alt: "open parenthesis 3 , 27 close parenthesis ,")[$( 3 , 27 ) ,$] for example. So this function is not one-to-one.

ⓑ  #linebreak() 
#math.equation(block: false, alt: "{ open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 9 , 3 close parenthesis , open parenthesis 16 , 4 close parenthesis }")[$&  &  &  & #h(5em) \{ ( 0 , 0 ) , ( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) , ( 16 , 4 ) \}$]

Each #emph[x]-value is matched with only one #emph[y]-value. So this relation is a function.

Since each #emph[y]-value is paired with only one #emph[x]-value, this function is one-to-one.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ #math.equation(block: false, alt: "{ open parenthesis −3 , −6 close parenthesis , open parenthesis −2 , −4 close parenthesis , open parenthesis −1 , −2 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 2 close parenthesis , open parenthesis 2 , 4 close parenthesis , open parenthesis 3 , 6 close parenthesis }")[$\{ ( −3 , −6 ) , ( −2 , −4 ) , ( −1 , −2 ) , ( 0 , 0 ) , ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) \}$] ⓑ #math.equation(block: false, alt: "{ open parenthesis −4 , 8 close parenthesis , open parenthesis −2 , 4 close parenthesis , open parenthesis −1 , 2 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 2 close parenthesis , open parenthesis 2 , 4 close parenthesis , open parenthesis 4 , 8 close parenthesis }")[$\{ ( −4 , 8 ) , ( −2 , 4 ) , ( −1 , 2 ) , ( 0 , 0 ) , ( 1 , 2 ) , ( 2 , 4 ) , ( 4 , 8 ) \}$]

#solutionbox[
ⓐ One-to-one function #linebreak() ⓑ Function; not one-to-one
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
For each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one.

ⓐ #math.equation(block: false, alt: "{ open parenthesis 27 , −3 close parenthesis , open parenthesis 8 , −2 close parenthesis , open parenthesis 1 , −1 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 8 , 2 close parenthesis , open parenthesis 27 , 3 close parenthesis }")[$\{ ( 27 , −3 ) , ( 8 , −2 ) , ( 1 , −1 ) , ( 0 , 0 ) , ( 1 , 1 ) , ( 8 , 2 ) , ( 27 , 3 ) \}$] ⓑ #math.equation(block: false, alt: "{ open parenthesis 7 , −3 close parenthesis , open parenthesis −5 , −4 close parenthesis , open parenthesis 8 , 0 close parenthesis , open parenthesis 0 , 0 close parenthesis , open parenthesis −6 , 4 close parenthesis , open parenthesis −2 , 2 close parenthesis , open parenthesis −1 , 3 close parenthesis }")[$\{ ( 7 , −3 ) , ( −5 , −4 ) , ( 8 , 0 ) , ( 0 , 0 ) , ( −6 , 4 ) , ( −2 , 2 ) , ( −1 , 3 ) \}$]

#solutionbox[
ⓐ Not a function #linebreak() ⓑ Function; not one-to-one
]
]

To help us determine whether a relation is a function, we use the #strong[vertical line test]. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. Also, if any vertical line intersects the graph in more than one point, the graph does not represent a function.

The vertical line is representing an #emph[x]-value and we check that it intersects the graph in only one #emph[y]-value. Then it is a function.

To check if a function is one-to-one, we use a similar process. We use a horizontal line and check that each horizontal line intersects the graph in only one point. The horizontal line is representing a #emph[y]-value and we check that it intersects the graph in only one #emph[x]-value. If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function. This is the #strong[horizontal line test].

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Horizontal Line Test]

If every horizontal line intersects the graph of a function in at most one point, it is a one-to-one function.
]

We can test whether a graph of a relation is a function by using the vertical line test. We can then tell if the function is one-to-one by applying the horizontal line test.

#examplebox("Example 4")[][
Determine ⓐ whether each graph is the graph of a function and, if so, ⓑ whether it is one-to-one.

#figure(figph[This first graph shows a straight line passing through (0, 2) and (3, 0). This second shows a parabola opening up with vertex at (0, negative 1).], alt: "This first graph shows a straight line passing through (0, 2) and (3, 0). This second shows a parabola opening up with vertex at (0, negative 1).", caption: none)

#solutionbox[
ⓐ  #linebreak() 
#figure(figph[This figure shows a straight line passing through (0, 2) and (3, 0), with a red vertical line that only passes through one point and a blue horizontal line that only passes through one point.], alt: "This figure shows a straight line passing through (0, 2) and (3, 0), with a red vertical line that only passes through one point and a blue horizontal line that only passes through one point.", caption: none)

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. Since any horizontal line intersects the graph in at most one point, the graph is the graph of a one-to-one function.

ⓑ  #linebreak() 
#figure(figph[This figure shows a parabola opening up with vertex at (0, negative 1), with a red vertical line that only passes through one point and a blue horizontal line that passes through two points.], alt: "This figure shows a parabola opening up with vertex at (0, negative 1), with a red vertical line that only passes through one point and a blue horizontal line that passes through two points.", caption: none)

Since any vertical line intersects the graph in at most one point, the graph is the graph of a function. The horizontal line shown on the graph intersects it in two points. This graph does not represent a one-to-one function.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

#figure(figph[Graph a shows a parabola opening to the right with vertex at (negative 1, 0). Graph b shows an exponential function that does not cross the x axis and that passes through (0, 1) before increasing rapidly.], alt: "Graph a shows a parabola opening to the right with vertex at (negative 1, 0). Graph b shows an exponential function that does not cross the x axis and that passes through (0, 1) before increasing rapidly.", caption: none)

#solutionbox[
ⓐ Not a function ⓑ One-to-one function
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Determine whether each graph is the graph of a function and, if so, whether it is one-to-one.

#figure(figph[Graph a shows a parabola opening up with vertex at (0, 3). Graph b shows a straight line passing through (0, negative 2) and (2, 0).], alt: "Graph a shows a parabola opening up with vertex at (0, 3). Graph b shows a straight line passing through (0, negative 2) and (2, 0).", caption: none)

#solutionbox[
ⓐ Function; not one-to-one ⓑ One-to-one function
]
]


=== Find the Inverse of a Function

Let’s look at a one-to one function, #math.equation(block: false, alt: "f")[$f$], represented by the ordered pairs #math.equation(block: false, alt: "{ open parenthesis 0 , 5 close parenthesis , open parenthesis 1 , 6 close parenthesis , open parenthesis 2 , 7 close parenthesis , open parenthesis 3 , 8 close parenthesis } .")[$\{ ( 0 , 5 ) , ( 1 , 6 ) , ( 2 , 7 ) , ( 3 , 8 ) \} .$] For each #math.equation(block: false, alt: "x")[$x$]-value, #math.equation(block: false, alt: "f")[$f$] adds 5 to get the #math.equation(block: false, alt: "y")[$y$]-value. To ‘undo’ the addition of 5, we subtract 5 from each #math.equation(block: false, alt: "y")[$y$]-value and get back to the original #math.equation(block: false, alt: "x")[$x$]-value. We can call this “taking the inverse of #math.equation(block: false, alt: "f")[$f$]” and name the function #math.equation(block: false, alt: "f to the power −1 .")[$f^(−1) .$]

#figure(figph[This figure shows the set (0, 5), (1, 6), (2, 7) and (3, 8) on the left side of an oval. The oval contains the numbers 0, 1, 2, and 3. There are black arrows from these numbers that point to the numbers 5, 6, 7, and 8, respectively in a second oval to the right of the first. Above this, there is a black arrow labeled “f add 5” coming from the left oval to the right oval. There are red arrows from the numbers 5, 6, 7, and 8 in the right oval to the numbers 0, 1, 2, and 3, respectively, in the left oval. Below this, we have a red arrow labeled “f with a superscript negative 1” and “subtract 5”. To the right of this, we have the set (5, 0), (6, 1), (7, 2) and (8, 3).], alt: "This figure shows the set (0, 5), (1, 6), (2, 7) and (3, 8) on the left side of an oval. The oval contains the numbers 0, 1, 2, and 3. There are black arrows from these numbers that point to the numbers 5, 6, 7, and 8, respectively in a second oval to the right of the first. Above this, there is a black arrow labeled “f add 5” coming from the left oval to the right oval. There are red arrows from the numbers 5, 6, 7, and 8 in the right oval to the numbers 0, 1, 2, and 3, respectively, in the left oval. Below this, we have a red arrow labeled “f with a superscript negative 1” and “subtract 5”. To the right of this, we have the set (5, 0), (6, 1), (7, 2) and (8, 3).", caption: none)

Notice that that the ordered pairs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power −1")[$f^(−1)$] have their #math.equation(block: false, alt: "x")[$x$]-values and #math.equation(block: false, alt: "y")[$y$]-values reversed. The domain of #math.equation(block: false, alt: "f")[$f$] is the range of #math.equation(block: false, alt: "f to the power −1")[$f^(−1)$] and the domain of #math.equation(block: false, alt: "f to the power −1")[$f^(−1)$] is the range of #math.equation(block: false, alt: "f .")[$f .$]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Inverse of a Function Defined by Ordered Pairs]

If #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a one-to-one function whose ordered pairs are of the form #math.equation(block: false, alt: "open parenthesis x , y close parenthesis ,")[$( x , y ) ,$] then its inverse function #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] is the set of ordered pairs #math.equation(block: false, alt: "open parenthesis y , x close parenthesis .")[$( y , x ) .$]
]

In the next example we will find the inverse of a function defined by ordered pairs.

#examplebox("Example 5")[][
Find the inverse of the function #math.equation(block: false, alt: "{ open parenthesis 0 , 3 close parenthesis , open parenthesis 1 , 5 close parenthesis , open parenthesis 2 , 7 close parenthesis , open parenthesis 3 , 9 close parenthesis } .")[$\{ ( 0 , 3 ) , ( 1 , 5 ) , ( 2 , 7 ) , ( 3 , 9 ) \} .$] Determine the domain and range of the inverse function.

#solutionbox[
This function is one-to-one since every #math.equation(block: false, alt: "x")[$x$]-value is paired with exactly one #math.equation(block: false, alt: "y")[$y$]-value.

To find the inverse we reverse the #math.equation(block: false, alt: "x")[$x$]-values and #math.equation(block: false, alt: "y")[$y$]-values in the ordered pairs of the function. #linebreak()

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([Function], [#math.equation(block: false, alt: "{ open parenthesis 0 , 3 close parenthesis , open parenthesis 1 , 5 close parenthesis , open parenthesis 2 , 7 close parenthesis , open parenthesis 3 , 9 close parenthesis }")[$\{ ( 0 , 3 ) , ( 1 , 5 ) , ( 2 , 7 ) , ( 3 , 9 ) \}$]]),
  [Inverse Function], [#math.equation(block: false, alt: "{ open parenthesis 3 , 0 close parenthesis , open parenthesis 5 , 1 close parenthesis , open parenthesis 7 , 2 close parenthesis , open parenthesis 9 , 3 close parenthesis }")[$\{ ( 3 , 0 ) , ( 5 , 1 ) , ( 7 , 2 ) , ( 9 , 3 ) \}$]],
  [Domain of Inverse Function], [#math.equation(block: false, alt: "{ 3 , 5 , 7 , 9 }")[$\{ 3 , 5 , 7 , 9 \}$]],
  [Range of Inverse Function], [#math.equation(block: false, alt: "{ 0 , 1 , 2 , 3 }")[$\{ 0 , 1 , 2 , 3 \}$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of #math.equation(block: false, alt: "{ open parenthesis 0 , 4 close parenthesis , open parenthesis 1 , 7 close parenthesis , open parenthesis 2 , 10 close parenthesis , open parenthesis 3 , 13 close parenthesis } .")[$\{ ( 0 , 4 ) , ( 1 , 7 ) , ( 2 , 10 ) , ( 3 , 13 ) \} .$] Determine the domain and range of the inverse function.

#solutionbox[
Inverse function: #math.equation(block: false, alt: "{ open parenthesis 4 , 0 close parenthesis , open parenthesis 7 , 1 close parenthesis , open parenthesis 10 , 2 close parenthesis , open parenthesis 13 , 3 close parenthesis } .")[$\{ ( 4 , 0 ) , ( 7 , 1 ) , ( 10 , 2 ) , ( 13 , 3 ) \} .$] Domain: #math.equation(block: false, alt: "{ 4 , 7 , 10 , 13 } .")[$\{ 4 , 7 , 10 , 13 \} .$] Range: #math.equation(block: false, alt: "{ 0 , 1 , 2 , 3 } .")[$\{ 0 , 1 , 2 , 3 \} .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of #math.equation(block: false, alt: "{ open parenthesis −1 , 4 close parenthesis , open parenthesis −2 , 1 close parenthesis , open parenthesis −3 , 0 close parenthesis , open parenthesis −4 , 2 close parenthesis } .")[$\{ ( −1 , 4 ) , ( −2 , 1 ) , ( −3 , 0 ) , ( −4 , 2 ) \} .$] Determine the domain and range of the inverse function.

#solutionbox[
Inverse function: #math.equation(block: false, alt: "{ open parenthesis 4 , −1 close parenthesis , open parenthesis 1 , −2 close parenthesis , open parenthesis 0 , −3 close parenthesis , open parenthesis 2 , −4 close parenthesis } .")[$\{ ( 4 , −1 ) , ( 1 , −2 ) , ( 0 , −3 ) , ( 2 , −4 ) \} .$] Domain: #math.equation(block: false, alt: "{ 0 , 1 , 2 , 4 } .")[$\{ 0 , 1 , 2 , 4 \} .$] Range: #math.equation(block: false, alt: "{ −4 , −3 , −2 , −1 } .")[$\{ −4 , −3 , −2 , −1 \} .$]
]
]

We just noted that if #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a one-to-one function whose ordered pairs are of the form #math.equation(block: false, alt: "open parenthesis x , y close parenthesis ,")[$( x , y ) ,$] then its inverse function #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] is the set of ordered pairs #math.equation(block: false, alt: "open parenthesis y , x close parenthesis .")[$( y , x ) .$]

So if a point #math.equation(block: false, alt: "open parenthesis a , b close parenthesis")[$( a , b )$] is on the graph of a function #math.equation(block: false, alt: "f open parenthesis x close parenthesis ,")[$f ( x ) ,$] then the ordered pair #math.equation(block: false, alt: "open parenthesis b , a close parenthesis")[$( b , a )$] is on the graph of #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis .")[$f^(−1) ( x ) .$]

#figure(figph[This figure shows the line y equals x with points (3,1) and (1,3) on either side of the line. These two points are connected by a dashed blue line segment.], alt: "This figure shows the line y equals x with points (3,1) and (1,3) on either side of the line. These two points are connected by a dashed blue line segment.", caption: none)

The distance between any two pairs #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")[$( b , a )$] is cut in half by the line #math.equation(block: false, alt: "y equals x .")[$y = x .$] So we say the points are mirror images of each other through the line #math.equation(block: false, alt: "y equals x .")[$y = x .$]

Since every point on the graph of a function #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a mirror image of a point on the graph of #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis ,")[$f^(−1) ( x ) ,$] we say the graphs are mirror images of each other through the line #math.equation(block: false, alt: "y equals x .")[$y = x .$] We will use this concept to graph the inverse of a function in the next example.

#examplebox("Example 6")[][
Graph, on the same coordinate system, the inverse of the one-to one function shown.

#figure(figph[This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1,0) then to (0,2) and then to (3, 4).], alt: "This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1,0) then to (0,2) and then to (3, 4).", caption: none)

#solutionbox[
We can use points on the graph to find points on the inverse graph. Some points on the graph are: #math.equation(block: false, alt: "open parenthesis −5 , −3 close parenthesis , open parenthesis −3 , −1 close parenthesis , open parenthesis −1 , 0 close parenthesis , open parenthesis 0 , 2 close parenthesis , open parenthesis 3 , 4 close parenthesis")[$( −5 , −3 ) , ( −3 , −1 ) , ( −1 , 0 ) , ( 0 , 2 ) , ( 3 , 4 )$].

So, the inverse function will contain the points: #math.equation(block: false, alt: "open parenthesis −3 , −5 close parenthesis , open parenthesis −1 , −3 close parenthesis , open parenthesis 0 , −1 close parenthesis , open parenthesis 2 , 0 close parenthesis , open parenthesis 4 , 3 close parenthesis")[$( −3 , −5 ) , ( −1 , −3 ) , ( 0 , −1 ) , ( 2 , 0 ) , ( 4 , 3 )$]. #linebreak() 
#figure(figph[This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1, 0) then to (0,2) and then to (3, 4). Then there is a dashed line to denote y equals x. There is also a line from (negative 3, negative 5) to (negative 1, negative 3) then to (0, negative 1), then to (2, 0) and then to (4, 3).], alt: "This figure shows a line from (negative 5, negative 3) to (negative 3, negative 1) then to (negative 1, 0) then to (0,2) and then to (3, 4). Then there is a dashed line to denote y equals x. There is also a line from (negative 3, negative 5) to (negative 1, negative 3) then to (0, negative 1), then to (2, 0) and then to (4, 3).", caption: none)

Notice how the graph of the original function and the graph of the inverse functions are mirror images through the line #math.equation(block: false, alt: "y equals x .")[$y = x .$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph, on the same coordinate system, the inverse of the one-to one function.

#figure(figph[The graph shows a line from (negative 3, negative 4) to (negative 2, negative 2) then to (0, negative 1), then to (1, 2) and then to (4, 3). The graph shows a line from (negative 3, 4) to (0, 3) then to (1, 2) and then to (4, 1).], alt: "The graph shows a line from (negative 3, negative 4) to (negative 2, negative 2) then to (0, negative 1), then to (1, 2) and then to (4, 3). The graph shows a line from (negative 3, 4) to (0, 3) then to (1, 2) and then to (4, 1).", caption: none)

#solutionbox[
#figure(figph[This figure shows a line from (negative 4, negative 3) to (negative 2, negative 2) then to (negative 1, 0) then to (2, 1) and then to (3, 4).], alt: "This figure shows a line from (negative 4, negative 3) to (negative 2, negative 2) then to (negative 1, 0) then to (2, 1) and then to (3, 4).", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Graph, on the same coordinate system, the inverse of the one-to one function.

#figure(figph[A graph on a Cartesian coordinate plane shows a piecewise linear function connecting four points: (-3, 4), (0, 3), (1, 2), and (4, 1).], alt: "A graph on a Cartesian coordinate plane shows a piecewise linear function connecting four points: (-3, 4), (0, 3), (1, 2), and (4, 1).", caption: none)

#solutionbox[
#figure(figph[Graph extends from negative 4 to 4 on both axes. Points plotted are (negative 3, 4), (0, 3), (1, 2), and (4, 1). Line segments connect points.], alt: "Graph extends from negative 4 to 4 on both axes. Points plotted are (negative 3, 4), (0, 3), (1, 2), and (4, 1). Line segments connect points.", caption: none)
]
]

When we began our discussion of an inverse function, we talked about how the inverse function ‘undoes’ what the original function did to a value in its domain in order to get back to the original #emph[x]-value.

#figure(figph[This figure shows x as the input to a box denoted as function f with f of x as the output of the box. Then, f of x is the input to a box denoted as function f superscript negative 1 with f superscript negative 1 of f of x equals x as the output of the box.], alt: "This figure shows x as the input to a box denoted as function f with f of x as the output of the box. Then, f of x is the input to a box denoted as function f superscript negative 1 with f superscript negative 1 of f of x equals x as the output of the box.", caption: none)

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Inverse Functions]

#math.equation(block: true, alt: "f to the power −1 open parenthesis f open parenthesis x close parenthesis close parenthesis, equals, x , for all x in the domain of f; f open parenthesis f to the power −1 open parenthesis x close parenthesis close parenthesis, equals, x , for all x in the domain of f to the power −1")[$f^(−1) ( f ( x ) ) & = & x , #h(0.2em) "for all" #h(0.2em) x #h(0.2em) "in the domain of" #h(0.2em) f \ f ( f^(−1) ( x ) ) & = & x , #h(0.2em) "for all" #h(0.2em) x #h(0.2em) "in the domain of" #h(0.2em) f^(−1)$]
]

We can use this property to verify that two functions are inverses of each other.

#examplebox("Example 7")[][
Verify that #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 x minus 1")[$f ( x ) = 5 x − 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x plus 1 over 5")[$g ( x ) = frac(x + 1, 5)$] are inverse functions.

#solutionbox[
The functions are inverses of each other if #math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x")[$g ( f ( x ) ) = x$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x .")[$f ( g ( x ) ) = x .$]

#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#figure(figph[Checking if g and f are inverse functions with the expression g(f(x)) =? x.], alt: "Checking if g and f are inverse functions with the expression g(f(x)) =? x.", caption: none)]),
  [Substitute #math.equation(block: false, alt: "5 x minus 1")[$5 x − 1$] for #math.equation(block: false, alt: "f open parenthesis x close parenthesis .")[$f ( x ) .$]], [#figure(figph[The image shows a mathematical equation in black text on a white background: g(5x - 1) =? x. The question mark above the equals sign indicates that the equation is being posed as a question or hypothesis.], alt: "The image shows a mathematical equation in black text on a white background: g(5x - 1) =? x. The question mark above the equals sign indicates that the equation is being posed as a question or hypothesis.", caption: none)],
  [#figure(figph[A mathematical problem asks to find the expression for g(5x - 1) given the function g(x) = (x + 1)/5, with the variable 'x' highlighted in red in the original function definition.], alt: "A mathematical problem asks to find the expression for g(5x - 1) given the function g(x) = (x + 1)/5, with the variable 'x' highlighted in red in the original function definition.", caption: none)], [#figure(figph[A mathematical equation where the expression ((5x - 1) + 1) / 5 is being evaluated to see if it equals x, indicated by a question mark over the equals sign.], alt: "A mathematical equation where the expression ((5x - 1) + 1) / 5 is being evaluated to see if it equals x, indicated by a question mark over the equals sign.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation shows 5x divided by 5, followed by a question mark over an equals sign, and then the variable x, asking if 5x/5 equals x.], alt: "A mathematical equation shows 5x divided by 5, followed by a question mark over an equals sign, and then the variable x, asking if 5x/5 equals x.", caption: none)],
  [Simplify.], [#figure(figph[A basic principle of logic and mathematics, 'X=X', is shown with a checkmark, emphasizing its self-evident truth.], alt: "A basic principle of logic and mathematics, 'X=X', is shown with a checkmark, emphasizing its self-evident truth.", caption: none)],
  [], [],
  [], [#figure(figph[A mathematical expression showing the composition of two functions, f(g(x)), questioning if it equals x. This setup is used to determine if f and g are inverse functions.], alt: "A mathematical expression showing the composition of two functions, f(g(x)), questioning if it equals x. This setup is used to determine if f and g are inverse functions.", caption: none)],
  [Substitute #math.equation(block: false, alt: "the fraction x plus 1 over 5")[$frac(x + 1, 5)$] for #math.equation(block: false, alt: "g open parenthesis x close parenthesis .")[$g ( x ) .$]], [#figure(figph[A mathematical equation is displayed on a white background. It shows 'f(x+1)/5' equals 'x'.], alt: "A mathematical equation is displayed on a white background. It shows 'f(x+1)/5' equals 'x'.", caption: none)],
  [#figure(figph[A mathematical problem asks to find the value of the function f((x+1)/5) given that f(x) = 5x - 1. The input to the function f is shown as a fraction with 'x+1' in red in the numerator and '5' in red in the denominator.], alt: "A mathematical problem asks to find the value of the function f((x+1)/5) given that f(x) = 5x - 1. The input to the function f is shown as a fraction with 'x+1' in red in the numerator and '5' in red in the denominator.", caption: none)], [#figure(figph[The image displays the equation 5((x+1)/5) - 1 =? x. Upon simplification, the 5s cancel out, resulting in (x+1) - 1, which equals x. Thus, the statement x = x is true, making the equation valid.], alt: "The image displays the equation 5((x+1)/5) - 1 =? x. Upon simplification, the 5s cancel out, resulting in (x+1) - 1, which equals x. Thus, the statement x = x is true, making the equation valid.", caption: none)],
  [Simplify.], [#figure(figph[A mathematical equation 'x + 1 - 1 = x' is displayed, with a question mark placed above the equals sign, posing whether the equality is true or false.], alt: "A mathematical equation 'x + 1 - 1 = x' is displayed, with a question mark placed above the equals sign, posing whether the equality is true or false.", caption: none)],
  [Simplify.], [#figure(figph[The equation X = X with a checkmark, symbolizing a correct or self-evident statement.], alt: "The equation X = X with a checkmark, symbolizing a correct or self-evident statement.", caption: none)],
))

Since both #math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x")[$g ( f ( x ) ) = x$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x")[$f ( g ( x ) ) = x$] are true, the functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 x minus 1")[$f ( x ) = 5 x − 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x plus 1 over 5")[$g ( x ) = frac(x + 1, 5)$] are inverse functions. That is, they are inverses of each other.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify that the functions are inverse functions.

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x minus 3")[$f ( x ) = 4 x − 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x plus 3 over 4 .")[$g ( x ) = frac(x + 3, 4) .$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses.
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify that the functions are inverse functions.

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x plus 6")[$f ( x ) = 2 x + 6$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x minus 6 over 2 .")[$g ( x ) = frac(x − 6, 2) .$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses.
]
]

We have found inverses of function defined by ordered pairs and from a graph. We will now look at how to find an inverse using an algebraic equation. The method uses the idea that if #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a one-to-one function with ordered pairs #math.equation(block: false, alt: "open parenthesis x , y close parenthesis ,")[$( x , y ) ,$] then its inverse function #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] is the set of ordered pairs #math.equation(block: false, alt: "open parenthesis y , x close parenthesis .")[$( y , x ) .$]

If we reverse the #emph[x] and #emph[y] in the function and then solve for #emph[y], we get our #strong[inverse function].

#examplebox("Example 8")[How to Find the inverse of a One-to-One Function][
Find the inverse of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x plus 7 .")[$f ( x ) = 4 x + 7 .$]

#solutionbox[
#figure(figph[Step 1 is to substitute y for f of x. To do so, we replace f of x with y. Hence, f of x equals 4 x plus 7 becomes y equals 4 x plus 7.], alt: "Step 1 is to substitute y for f of x. To do so, we replace f of x with y. Hence, f of x equals 4 x plus 7 becomes y equals 4 x plus 7.", caption: none)

#figure(figph[Step 2 is to interchange the variables x and y. To do so, we replace x with y and then y with x. Hence, we obtain x equals 4y plus 7.], alt: "Step 2 is to interchange the variables x and y. To do so, we replace x with y and then y with x. Hence, we obtain x equals 4y plus 7.", caption: none)

#figure(figph[Step 3 is to solve for y. To do so, we subtract 7 from each side and then divide by 4. Hence, we have x minus 7 equals 4y and then the quantity x minus 7 divided by 4 equals y.], alt: "Step 3 is to solve for y. To do so, we subtract 7 from each side and then divide by 4. Hence, we have x minus 7 equals 4y and then the quantity x minus 7 divided by 4 equals y.", caption: none)

#figure(figph[Step 4 is to substitute f superscript negative 1 of x for y. To do so, we replace y with f superscript negative 1 of x. Hence, the quantity x minus 7 divided by 4 equals f superscript negative 1 of x.], alt: "Step 4 is to substitute f superscript negative 1 of x for y. To do so, we replace y with f superscript negative 1 of x. Hence, the quantity x minus 7 divided by 4 equals f superscript negative 1 of x.", caption: none)

#figure(figph[Step 5 is to verify that the functions are inverses. To do so, we show that f superscript negative 1 of f of x equals x and that f of f superscript negative 1of x equals x. Hence, we ask whether f inverse of 4x plus 7 equals x. This becomes a question of whether 4 x plus 7 minus 7 all divided by 4 equals x. This becomes a question of whether 4x divided by 4 equals x. This is true. To show the other side, we examine whether f of f inverse of x equals x. This becomes a question of whether f of the quantity x minus 7 divided by 4 equals x. This becomes a question of whether 4 times the quantity x minus 7 divided by 4 equals x. This becomes a question of whether x minus 7 plus 7 equals x. This is true.], alt: "Step 5 is to verify that the functions are inverses. To do so, we show that f superscript negative 1 of f of x equals x and that f of f superscript negative 1of x equals x. Hence, we ask whether f inverse of 4x plus 7 equals x. This becomes a question of whether 4 x plus 7 minus 7 all divided by 4 equals x. This becomes a question of whether 4x divided by 4 equals x. This is true. To show the other side, we examine whether f of f inverse of x equals x. This becomes a question of whether f of the quantity x minus 7 divided by 4 equals x. This becomes a question of whether 4 times the quantity x minus 7 divided by 4 equals x. This becomes a question of whether x minus 7 plus 7 equals x. This is true.", caption: none)
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 x minus 3 .")[$f ( x ) = 5 x − 3 .$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x plus 3 over 5")[$f^(−1) ( x ) = frac(x + 3, 5)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 8 x plus 5 .")[$f ( x ) = 8 x + 5 .$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x minus 5 over 8")[$f^(−1) ( x ) = frac(x − 5, 8)$]
]
]

We summarize the steps below.

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[How to Find the inverse of a One-to-One Function]

+ Substitute #emph[y] for #math.equation(block: false, alt: "f open parenthesis x close parenthesis .")[$f ( x ) .$]
+ Interchange the variables #emph[x] and #emph[y].
+ Solve for #emph[y].
+ Substitute #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] for #emph[y].
+ Verify that the functions are inverses.
]

#examplebox("Example 9")[How to Find the Inverse of a One-to-One Function][
Find the inverse of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the 5th root of 2 x minus 3 .")[$f ( x ) = root(5, 2 x − 3) .$]

#solutionbox[
#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the 5th root of 2 x minus 3")[$f ( x ) = root(5, 2 x − 3) #h(2.8em)$]]),
  [Substitute #math.equation(block: false, alt: "y")[$y$] for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].], [#math.equation(block: false, alt: "y equals the 5th root of 2 x minus 3")[$y = root(5, 2 x − 3) #h(2.8em)$]],
  [Interchange the variables #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$].], [#math.equation(block: false, alt: "x equals the 5th root of 2 y minus 3")[$x = root(5, 2 y − 3) #h(2.8em)$]],
  [Solve for #math.equation(block: false, alt: "y")[$y$].], [#math.equation(block: false, alt: "open parenthesis x close parenthesis to the power 5 equals open parenthesis the 5th root of 2 y minus 3 close parenthesis to the power 5")[$attach(( x ), t: 5) = attach(( root(5, 2 y − 3) ), t: 5) #h(1.3em)$]],
  [], [#math.equation(block: false, alt: "x to the power 5 equals 2 y minus 3")[$x^(5) = 2 y − 3 #h(3.1em)$]],
  [], [#math.equation(block: false, alt: "x to the power 5 plus 3 equals 2 y")[$x^(5) + 3 = 2 y #h(4.8em)$]],
  [], [#math.equation(block: false, alt: "the fraction x to the power 5 plus 3 over 2 equals y")[$frac(x^(5) + 3, 2) = y #h(5.4em)$]],
  [Substitute #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$] for #math.equation(block: false, alt: "y")[$y$].], [#math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction x to the power 5 plus 3 over 2")[$f^(− 1) ( x ) = frac(x^(5) + 3, 2) #h(3.4em)$]],
  [Verify that the functions are inverses.], [],
  [#math.equation(block: false, alt: "f to the power minus 1 open parenthesis f open parenthesis x close parenthesis close parenthesis equals ? x")[$f^(− 1) ( f ( x ) ) limits(=)^(?) x #h(4em)$]], [#math.equation(block: false, alt: "f open parenthesis f to the power minus 1 open parenthesis x close parenthesis close parenthesis equals ? x")[$f ( f^(− 1) ( x ) ) limits(=)^(?) x #h(5.4em)$]],
  [#math.equation(block: false, alt: "f to the power minus 1 open parenthesis the 5th root of 2 x minus 3 close parenthesis equals ? x")[$f^(− 1) ( root(5, 2 x − 3) ) limits(=)^(?) x #h(4em)$]], [#math.equation(block: false, alt: "f open parenthesis the fraction x to the power 5 plus 3 over 2 close parenthesis equals ? x")[$f ( frac(x^(5) + 3, 2) ) limits(=)^(?) x #h(5.4em)$]],
  [#math.equation(block: false, alt: "the fraction open parenthesis the 5th root of 2 x minus 3 close parenthesis to the power 5 plus 3 over 2 equals ? x")[$frac(attach(( root(5, 2 x − 3) ), t: 5) + 3, 2) limits(=)^(?) x #h(4em)$]], [#math.equation(block: false, alt: "the 5th root of 2 open parenthesis the fraction x to the power 5 plus 3 over 2 close parenthesis minus 3 equals ? x")[$root(5, 2 ( frac(x^(5) + 3, 2) ) − 3) limits(=)^(?) x #h(5.4em)$]],
  [#math.equation(block: false, alt: "the fraction 2 x minus 3 plus 3 over 2 equals ? x")[$frac(2 x − 3 + 3, 2) limits(=)^(?) x #h(4em)$]], [#math.equation(block: false, alt: "the 5th root of x to the power 5 plus 3 minus 3 equals ? x")[$root(5, x^(5) + 3 − 3) limits(=)^(?) x #h(5.4em)$]],
  [#math.equation(block: false, alt: "the fraction 2 x over 2 equals ? x")[$frac(2 x, 2) limits(=)^(?) x #h(4em)$]], [#math.equation(block: false, alt: "the 5th root of x to the power 5 equals ? x")[$root(5, x^(5)) limits(=)^(?) x #h(5.4em)$]],
  [#math.equation(block: false, alt: "x equals x ✓")[$x = x ✓ #h(3.2em)$]], [#math.equation(block: false, alt: "x equals x ✓")[$x = x ✓ #h(4.1em)$]],
))
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the 5th root of 3 x minus 2 .")[$f ( x ) = root(5, 3 x − 2) .$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x to the power 5 plus 2 over 3")[$f^(−1) ( x ) = frac(x^(5) + 2, 3)$]
]
]

#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the 4th root of 6 x minus 7 .")[$f ( x ) = root(4, 6 x − 7) .$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x to the power 4 plus 7 over 6")[$f^(−1) ( x ) = frac(x^(4) + 7, 6)$]
]
]


=== Key Concepts

- #strong[Composition of Functions:] The composition of functions #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g ,")[$g ,$] is written #math.equation(block: false, alt: "f ∘ g")[$f ∘ g$] and is defined by #linebreak() 
#math.equation(block: true, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis equals f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$( f ∘ g ) ( x ) = f ( g ( x ) )$] #linebreak() 
We read #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis")[$f ( g ( x ) )$] as #math.equation(block: false, alt: "f")[$f$] of #math.equation(block: false, alt: "g")[$g$] of #math.equation(block: false, alt: "x .")[$x .$]
- #strong[Horizontal Line Test:] If every horizontal line, intersects the graph of a function in at most one point, it is a one-to-one function.
- #strong[Inverse of a Function Defined by Ordered Pairs:] If #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is a one-to-one function whose ordered pairs are of the form #math.equation(block: false, alt: "open parenthesis x , y close parenthesis ,")[$( x , y ) ,$] then its inverse function #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] is the set of ordered pairs #math.equation(block: false, alt: "open parenthesis y , x close parenthesis .")[$( y , x ) .$]
- #strong[Inverse Functions:] For every #math.equation(block: false, alt: "x")[$x$] in the domain of one-to-one function #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power −1 ,")[$f^(−1) ,$] #linebreak() 
#math.equation(block: true, alt: "f to the power −1 open parenthesis f open parenthesis x close parenthesis close parenthesis, equals, x; f open parenthesis f to the power −1 open parenthesis x close parenthesis close parenthesis, equals, x")[$f^(−1) ( f ( x ) ) & = & x \ f ( f^(−1) ( x ) ) & = & x$]
- #strong[How to Find the Inverse of a One-to-One Function:]
+ Substitute #emph[y] for #math.equation(block: false, alt: "f open parenthesis x close parenthesis .")[$f ( x ) .$]
+ Interchange the variables #emph[x] and #emph[y].
+ Solve for #emph[y].
+ Substitute #math.equation(block: false, alt: "f to the power −1 open parenthesis x close parenthesis")[$f^(−1) ( x )$] for #math.equation(block: false, alt: "y .")[$y .$]
+ Verify that the functions are inverses.


==== Practice Makes Perfect

#strong[Find and Evaluate Composite Functions]

In the following exercises, find ⓐ (#emph[f] ∘ #emph[g])(#emph[x]), ⓑ (#emph[g] ∘ #emph[f])(#emph[x]), and ⓒ (#emph[f] · #emph[g])(#emph[x]).

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x plus 3")[$f ( x ) = 4 x + 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x plus 5")[$g ( x ) = 2 x + 5$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "8 x plus 23")[$8 x + 23$] ⓑ #math.equation(block: false, alt: "8 x plus 11")[$8 x + 11$] ⓒ  #linebreak() #math.equation(block: false, alt: "8 x squared plus 26 x plus 15")[$8 x^(2) + 26 x + 15$]
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x minus 1")[$f ( x ) = 3 x − 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 x minus 3")[$g ( x ) = 5 x − 3$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 6 x minus 5")[$f ( x ) = 6 x − 5$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 4 x plus 1")[$g ( x ) = 4 x + 1$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "24 x plus 1")[$24 x + 1$] ⓑ #math.equation(block: false, alt: "24 x minus 19")[$24 x − 19$] #linebreak() ⓒ #math.equation(block: false, alt: "24 x squared minus 14 x minus 5")[$24 x^(2) − 14 x − 5$]
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x plus 7")[$f ( x ) = 2 x + 7$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x minus 4")[$g ( x ) = 3 x − 4$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x")[$f ( x ) = 3 x$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x squared minus 3 x")[$g ( x ) = 2 x^(2) − 3 x$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "6 x squared minus 9 x")[$6 x^(2) − 9 x$] ⓑ #math.equation(block: false, alt: "18 x squared minus 9 x")[$18 x^(2) − 9 x$] #linebreak() ⓒ #math.equation(block: false, alt: "6 x cubed minus 9 x squared")[$6 x^(3) − 9 x^(2)$]
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x")[$f ( x ) = 2 x$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x squared minus 1")[$g ( x ) = 3 x^(2) − 1$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x minus 1")[$f ( x ) = 2 x − 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared plus 2")[$g ( x ) = x^(2) + 2$]

#solutionbox[
ⓐ #math.equation(block: false, alt: "2 x squared plus 3")[$2 x^(2) + 3$] ⓑ #math.equation(block: false, alt: "4 x squared minus 4 x plus 3")[$4 x^(2) − 4 x + 3$] #linebreak() ⓒ #math.equation(block: false, alt: "2 x cubed minus x squared plus 4 x minus 2")[$2 x^(3) − x^(2) + 4 x − 2$]
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x plus 3")[$f ( x ) = 4 x + 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared minus 4")[$g ( x ) = x^(2) − 4$]

In the following exercises, find the values described.

For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x squared plus 3")[$f ( x ) = 2 x^(2) + 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 x minus 1 ,")[$g ( x ) = 5 x − 1 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −2 close parenthesis")[$( f ∘ g ) ( −2 )$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −3 close parenthesis")[$( g ∘ f ) ( −3 )$] ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis −1 close parenthesis")[$( f ∘ f ) ( −1 )$]

#solutionbox[
ⓐ 245 ⓑ 104 ⓒ 53
]

For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 x squared minus 1")[$f ( x ) = 5 x^(2) − 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 4 x minus 1 ,")[$g ( x ) = 4 x − 1 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis 1 close parenthesis")[$( f ∘ g ) ( 1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −1 close parenthesis")[$( g ∘ f ) ( −1 )$] ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis 2 close parenthesis")[$( f ∘ f ) ( 2 )$]

For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x cubed")[$f ( x ) = 2 x^(3)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x squared plus 2 ,")[$g ( x ) = 3 x^(2) + 2 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −1 close parenthesis")[$( f ∘ g ) ( −1 )$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis 1 close parenthesis")[$( g ∘ f ) ( 1 )$] ⓒ #math.equation(block: false, alt: "open parenthesis g ∘ g close parenthesis open parenthesis 1 close parenthesis")[$( g ∘ g ) ( 1 )$]

#solutionbox[
ⓐ 250 ⓑ 14 ⓒ 77
]

For functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x cubed plus 1")[$f ( x ) = 3 x^(3) + 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x squared minus 3 ,")[$g ( x ) = 2 x^(2) − 3 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −2 close parenthesis")[$( f ∘ g ) ( −2 )$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −1 close parenthesis")[$( g ∘ f ) ( −1 )$] ⓒ #math.equation(block: false, alt: "open parenthesis g ∘ g close parenthesis open parenthesis 1 close parenthesis")[$( g ∘ g ) ( 1 )$]

#strong[Determine Whether a Function is One-to-One]

In the following exercises, determine if the set of ordered pairs represents a function and if so, is the function one-to-one.

#math.equation(block: true, alt: "{ open parenthesis −3 , 9 close parenthesis , open parenthesis −2 , 4 close parenthesis , open parenthesis −1 , 1 close parenthesis , open parenthesis 0 , 0 close parenthesis")[$\{ ( −3 , 9 ) , ( −2 , 4 ) , ( −1 , 1 ) , ( 0 , 0 )$], #linebreak() #math.equation(block: true, alt: "open parenthesis 1 , 1 close parenthesis , open parenthesis 2 , 4 close parenthesis , open parenthesis 3 , 9 close parenthesis }")[$( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) \}$]

#solutionbox[
Function; not one-to-one
]

#math.equation(block: true, alt: "{ open parenthesis 9 , −3 close parenthesis , open parenthesis 4 , −2 close parenthesis , open parenthesis 1 , −1 close parenthesis , open parenthesis 0 , 0 close parenthesis")[$\{ ( 9 , −3 ) , ( 4 , −2 ) , ( 1 , −1 ) , ( 0 , 0 )$], #linebreak() #math.equation(block: true, alt: "open parenthesis 1 , 1 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 9 , 3 close parenthesis }")[$( 1 , 1 ) , ( 4 , 2 ) , ( 9 , 3 ) \}$]

#math.equation(block: true, alt: "{ open parenthesis −3 , −5 close parenthesis , open parenthesis −2 , −3 close parenthesis , open parenthesis −1 , −1 close parenthesis")[$\{ ( −3 , −5 ) , ( −2 , −3 ) , ( −1 , −1 )$], #linebreak() #math.equation(block: true, alt: "open parenthesis 0 , 1 close parenthesis , open parenthesis 1 , 3 close parenthesis , open parenthesis 2 , 5 close parenthesis , open parenthesis 3 , 7 close parenthesis }")[$( 0 , 1 ) , ( 1 , 3 ) , ( 2 , 5 ) , ( 3 , 7 ) \}$]

#solutionbox[
One-to-one function
]

#math.equation(block: true, alt: "{ open parenthesis 5 , 3 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 3 , 1 close parenthesis , open parenthesis 2 , 0 close parenthesis")[$\{ ( 5 , 3 ) , ( 4 , 2 ) , ( 3 , 1 ) , ( 2 , 0 )$], #linebreak() #math.equation(block: true, alt: "open parenthesis 1 , −1 close parenthesis , open parenthesis 0 , −2 close parenthesis , open parenthesis −1 , −3 close parenthesis }")[$( 1 , −1 ) , ( 0 , −2 ) , ( −1 , −3 ) \}$]

In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one.

ⓐ  #linebreak() 
#figure(figph[This figure shows a graph of a circle with center at the origin and radius 3.], alt: "This figure shows a graph of a circle with center at the origin and radius 3.", caption: none)

 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows a graph of a parabola opening upward with vertex at (0k, 2).], alt: "This figure shows a graph of a parabola opening upward with vertex at (0k, 2).", caption: none)

#solutionbox[
ⓐ Not a function ⓑ Function; not one-to-one
]

ⓐ  #linebreak() 
#figure(figph[This figure shows a parabola opening to the right with vertex at (negative 2, 0).], alt: "This figure shows a parabola opening to the right with vertex at (negative 2, 0).", caption: none)

 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows a graph of a polynomial with odd order, so that it starts in the third quadrant, increases to the origin and then continues increasing through the first quadrant.], alt: "This figure shows a graph of a polynomial with odd order, so that it starts in the third quadrant, increases to the origin and then continues increasing through the first quadrant.", caption: none)

ⓐ  #linebreak() 
#figure(figph[This figure shows a graph of a curve that starts at (negative 6 negative 2) increases to the origin and then continues increasing slowly to (6, 2).], alt: "This figure shows a graph of a curve that starts at (negative 6 negative 2) increases to the origin and then continues increasing slowly to (6, 2).", caption: none)

 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows a parabola opening upward with vertex at (0, negative 4).], alt: "This figure shows a parabola opening upward with vertex at (0, negative 4).", caption: none)

#solutionbox[
ⓐ One-to-one function #linebreak() ⓑ Function; not one-to-one
]

ⓐ  #linebreak() 
#figure(figph[This figure shows a straight line segment decreasing from (negative 4, 6) to (2, 0), after which it increases from (2, 0) to (6, 4).], alt: "This figure shows a straight line segment decreasing from (negative 4, 6) to (2, 0), after which it increases from (2, 0) to (6, 4).", caption: none)

 #linebreak() 
ⓑ  #linebreak() 
#figure(figph[This figure shows a circle with radius 4 and center at the origin.], alt: "This figure shows a circle with radius 4 and center at the origin.", caption: none)

In the following exercises, find the inverse of each function. Determine the domain and range of the inverse function.

#math.equation(block: true, alt: "{ open parenthesis 2 , 1 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 6 , 3 close parenthesis , open parenthesis 8 , 4 close parenthesis }")[$\{ ( 2 , 1 ) , ( 4 , 2 ) , ( 6 , 3 ) , ( 8 , 4 ) \}$]

#solutionbox[
Inverse function: #math.equation(block: false, alt: "{ open parenthesis 1 , 2 close parenthesis , open parenthesis 2 , 4 close parenthesis , open parenthesis 3 , 6 close parenthesis , open parenthesis 4 , 8 close parenthesis } .")[$\{ ( 1 , 2 ) , ( 2 , 4 ) , ( 3 , 6 ) , ( 4 , 8 ) \} .$] Domain: #math.equation(block: false, alt: "{ 1 , 2 , 3 , 4 } .")[$\{ 1 , 2 , 3 , 4 \} .$] Range: #math.equation(block: false, alt: "{ 2 , 4 , 6 , 8 } .")[$\{ 2 , 4 , 6 , 8 \} .$]
]

#math.equation(block: true, alt: "{ open parenthesis 6 , 2 close parenthesis , open parenthesis 9 , 5 close parenthesis , open parenthesis 12 , 8 close parenthesis , open parenthesis 15 , 11 close parenthesis }")[$\{ ( 6 , 2 ) , ( 9 , 5 ) , ( 12 , 8 ) , ( 15 , 11 ) \}$]

#math.equation(block: true, alt: "{ open parenthesis 0 , −2 close parenthesis , open parenthesis 1 , 3 close parenthesis , open parenthesis 2 , 7 close parenthesis , open parenthesis 3 , 12 close parenthesis }")[$\{ ( 0 , −2 ) , ( 1 , 3 ) , ( 2 , 7 ) , ( 3 , 12 ) \}$]

#solutionbox[
Inverse function: #math.equation(block: false, alt: "{ open parenthesis −2 , 0 close parenthesis , open parenthesis 3 , 1 close parenthesis , open parenthesis 7 , 2 close parenthesis , open parenthesis 12 , 3 close parenthesis } .")[$\{ ( −2 , 0 ) , ( 3 , 1 ) , ( 7 , 2 ) , ( 12 , 3 ) \} .$] Domain: #math.equation(block: false, alt: "{ −2 , 3 , 7 , 12 } .")[$\{ −2 , 3 , 7 , 12 \} .$] Range: #math.equation(block: false, alt: "{ 0 , 1 , 2 , 3 } .")[$\{ 0 , 1 , 2 , 3 \} .$]
]

#math.equation(block: true, alt: "{ open parenthesis 0 , 0 close parenthesis , open parenthesis 1 , 1 close parenthesis , open parenthesis 2 , 4 close parenthesis , open parenthesis 3 , 9 close parenthesis }")[$\{ ( 0 , 0 ) , ( 1 , 1 ) , ( 2 , 4 ) , ( 3 , 9 ) \}$]

#math.equation(block: true, alt: "{ open parenthesis −2 , −3 close parenthesis , open parenthesis −1 , −1 close parenthesis , open parenthesis 0 , 1 close parenthesis , open parenthesis 1 , 3 close parenthesis }")[$\{ ( −2 , −3 ) , ( −1 , −1 ) , ( 0 , 1 ) , ( 1 , 3 ) \}$]

#solutionbox[
Inverse function: #math.equation(block: false, alt: "{ open parenthesis −3 , − 2 close parenthesis , open parenthesis −1 , −1 close parenthesis , open parenthesis 1 , 0 close parenthesis , open parenthesis 3 , 1 close parenthesis } .")[$\{ ( −3 , "−" 2 ) , ( −1 , −1 ) , ( 1 , 0 ) , ( 3 , 1 ) \} .$] Domain: #math.equation(block: false, alt: "{ −3 , − 1 , 1 , 3 } .")[$\{ −3 , "−" 1 , 1 , 3 \} .$] Range: #math.equation(block: false, alt: "{ −2 , −1 , 0 , 1 } .")[$\{ −2 , −1 , 0 , 1 \} .$]
]

#math.equation(block: true, alt: "{ open parenthesis 5 , 3 close parenthesis , open parenthesis 4 , 2 close parenthesis , open parenthesis 3 , 1 close parenthesis , open parenthesis 2 , 0 close parenthesis }")[$\{ ( 5 , 3 ) , ( 4 , 2 ) , ( 3 , 1 ) , ( 2 , 0 ) \}$]

In the following exercises, graph, on the same coordinate system, the inverse of the one-to-one function shown.

#figure(figph[This figure shows a series of line segments from (negative 4, negative 3) to (negative 3, 0) then to (negative 1, 2) and then to (3, 4).], alt: "This figure shows a series of line segments from (negative 4, negative 3) to (negative 3, 0) then to (negative 1, 2) and then to (3, 4).", caption: none)

#solutionbox[
#figure(figph[This figure shows a series of line segments from (negative 3, negative 4) to (0, negative 3) then to (2, negative 1), and then to (4, 3).], alt: "This figure shows a series of line segments from (negative 3, negative 4) to (0, negative 3) then to (2, negative 1), and then to (4, 3).", caption: none)
]

#figure(figph[This figure shows a series of line segments from (negative 4, negative 4) to (negative 3, 1) then to (0, 2) and then to (2, 4).], alt: "This figure shows a series of line segments from (negative 4, negative 4) to (negative 3, 1) then to (0, 2) and then to (2, 4).", caption: none)

#figure(figph[This figure shows a series of line segments from (negative 4, 4) to (0, 3) then to (3, 2) and then to (4, negative 1).], alt: "This figure shows a series of line segments from (negative 4, 4) to (0, 3) then to (3, 2) and then to (4, negative 1).", caption: none)

#solutionbox[
#figure(figph[This figure shows a series of line segments from (negative 1, 4) to (2, 3) then to (3, 0), and then to (4, negative 4).], alt: "This figure shows a series of line segments from (negative 1, 4) to (2, 3) then to (3, 0), and then to (4, negative 4).", caption: none)
]

#figure(figph[This figure shows a series of line segments from (negative 4, negative 4) to (negative 1, negative 3) then to (0, 1), then to (1, 3), and then to (4, 4).], alt: "This figure shows a series of line segments from (negative 4, negative 4) to (negative 1, negative 3) then to (0, 1), then to (1, 3), and then to (4, 4).", caption: none)

In the following exercises, determine whether or not the given functions are inverses.

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x plus 8")[$f ( x ) = x + 8$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x minus 8")[$g ( x ) = x − 8$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses.
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x minus 9")[$f ( x ) = x − 9$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x plus 9")[$g ( x ) = x + 9$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 7 x")[$f ( x ) = 7 x$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x over 7")[$g ( x ) = frac(x, 7)$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses.
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction x over 11")[$f ( x ) = frac(x, 11)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 11 x")[$g ( x ) = 11 x$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 7 x plus 3")[$f ( x ) = 7 x + 3$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x minus 3 over 7")[$g ( x ) = frac(x − 3, 7)$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses.
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 x minus 4")[$f ( x ) = 5 x − 4$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x minus 4 over 5")[$g ( x ) = frac(x − 4, 5)$]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the square root of x plus 2")[$f ( x ) = sqrt(x + 2)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared minus 2 open parenthesis x greater than 0 close parenthesis")[$g ( x ) = x^(2) − 2 ( x > 0 )$]

#solutionbox[
#math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses (for nonnegative #math.equation(block: false, alt: "x close parenthesis .")[$x \) .$]
]

#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the cube root of x minus 4")[$f ( x ) = root(3, x − 4)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x cubed plus 4")[$g ( x ) = x^(3) + 4$]

In the following exercises, find the inverse of each function.

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x minus 12")[$f ( x ) = x − 12$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals x plus 12")[$f^(−1) ( x ) = x + 12$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x plus 17")[$f ( x ) = x + 17$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 9 x")[$f ( x ) = 9 x$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x over 9")[$f^(−1) ( x ) = frac(x, 9)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 8 x")[$f ( x ) = 8 x$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction x over 6")[$f ( x ) = frac(x, 6)$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals 6 x")[$f^(−1) ( x ) = 6 x$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction x over 4")[$f ( x ) = frac(x, 4)$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 x minus 7")[$f ( x ) = 6 x − 7$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x plus 7 over 6")[$f^(−1) ( x ) = frac(x + 7, 6)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 7 x minus 1")[$f ( x ) = 7 x − 1$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals −2 x plus 5")[$f ( x ) = −2 x + 5$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x minus 5 over −2")[$f^(−1) ( x ) = frac(x − 5, −2)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals −5 x minus 4")[$f ( x ) = −5 x − 4$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared plus 6 ,")[$f ( x ) = x^(2) + 6 ,$] #math.equation(block: true, alt: "x greater than or equal to 0")[$x ≥ 0$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the square root of x minus 6")[$f^(−1) ( x ) = sqrt(x − 6)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 9 ,")[$f ( x ) = x^(2) − 9 ,$] #math.equation(block: true, alt: "x greater than or equal to 0")[$x ≥ 0$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed minus 4")[$f ( x ) = x^(3) − 4$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the cube root of x plus 4")[$f^(−1) ( x ) = root(3, x + 4)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed plus 6")[$f ( x ) = x^(3) + 6$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x plus 2")[$f ( x ) = frac(1, x + 2)$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction 1 over x minus 2")[$f^(−1) ( x ) = frac(1, x) − 2$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x minus 6")[$f ( x ) = frac(1, x − 6)$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the square root of x minus 2 ,")[$f ( x ) = sqrt(x − 2) ,$] #math.equation(block: true, alt: "x greater than or equal to 2")[$x ≥ 2$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals x squared plus 2")[$f^(−1) ( x ) = x^(2) + 2$], #math.equation(block: true, alt: "x greater than or equal to 0")[$x ≥ 0$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the square root of x plus 8 ,")[$f ( x ) = sqrt(x + 8) ,$] #math.equation(block: true, alt: "x greater than or equal to −8")[$x ≥ −8$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the cube root of x minus 3")[$f ( x ) = root(3, x − 3)$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals x cubed plus 3")[$f^(−1) ( x ) = x^(3) + 3$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the cube root of x plus 5")[$f ( x ) = root(3, x + 5)$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the 4th root of 9 x minus 5 ,")[$f ( x ) = root(4, 9 x − 5) ,$] #math.equation(block: true, alt: "x greater than or equal to the fraction 5 over 9")[$x ≥ frac(5, 9)$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x to the power 4 plus 5 over 9")[$f^(−1) ( x ) = frac(x^(4) + 5, 9)$], #math.equation(block: true, alt: "x greater than or equal to 0")[$x ≥ 0$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the 4th root of 8 x minus 3 ,")[$f ( x ) = root(4, 8 x − 3) ,$] #math.equation(block: true, alt: "x greater than or equal to the fraction 3 over 8")[$x ≥ frac(3, 8)$]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the 5th root of −3 x plus 5")[$f ( x ) = root(5, −3 x + 5)$]

#solutionbox[
#math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x to the power 5 minus 5 over −3")[$f^(−1) ( x ) = frac(x^(5) − 5, −3)$]
]

#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the 5th root of −4 x minus 3")[$f ( x ) = root(5, −4 x − 3)$]


==== Writing Exercises

Explain how the graph of the inverse of a function is related to the graph of the function.

#solutionbox[
Answers will vary.
]

Explain how to find the inverse of a function from its equation. Use an example to demonstrate the steps.


==== Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

#figure(figph[This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Find and evaluate composite functions, determine whether a function is one-to-one, and find the inverse of a function. The rest of the cells are blank.], alt: "This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads Find and evaluate composite functions, determine whether a function is one-to-one, and find the inverse of a function. The rest of the cells are blank.", caption: none)

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.