5.3 Function Composition

Related Rates

In Section 5.2.1, we studied the difference operator, $\mathrm{\Delta }$ and showed how average rates of change operate with the basic function arithmetic. In this section, we explore how rates of change of composite functions are related to the rates of change of their constituent functions. As in that section, we'll use the formulation:

$$\frac{\mathrm{\Delta }[f(x)]}{\mathrm{\Delta }x}=\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x},\mathrm{\Delta }x\ne 0,$$

adjusting the names of functions and independent variables as needed.

As a motivational example, we revisit the scenario in Example 5.3.4.

Example 5.3.5 verifies a property of rates we formalize below.

If we think of $u$ as being an `intermediary' variable, Theorem 5.5 allows us to determine the rate of change of $f$ with respect to $x$ by multiplying the rate of change of $f$ with respect to this `intermediary' $u$ by the rate of change of the `intermediary' $u$ with respect to $x$. That is, we are looking for rate information on how $f$ depends on $x$ by decomposing the rate into two rates as visualized below.

We close the section with one last example.

Note that we never needed to know explicitly how the speed, $v$, directly depended on time in order to answer the question posed in Example 5.3.6. All we needed was the rate.

Exercises

In Exercises -, use the given pair of functions to find the following values if they exist.

• $(g\circ f)(0)$
• $(f\circ g)(-1)$
• $(f\circ f)(2)$

• $(g\circ f)(-3)$
• $(f\circ g)\left(\frac{1}{2}\right)$
• $(f\circ f)(-2)$

1. $f(x)=x^2$, $g(t)=2t+1$
2. $f(x)=4-x$, $g(t)=1-t^2$
3. $f(x)=4-3x$, $g(t)=|t|$
4. $f(x)=|x-1|$, $g(t)=t^2-5$
5. $f(x)=4x+5$, $g(t)=\sqrt{t}$
6. $f(x)=\sqrt{3-x}$, $g(t)=t^2+1$
7. $f(x)=6-x-x^2$, $g(t)=t\sqrt{t+10}$
8. $f(x)=\sqrt[3]{x+1}$, $g(t)=4t^2-t$
9. $f(x)=\frac{3}{1-x}$, $g(t)=\frac{4t}{t^2+1}$
10. $f(x)=\frac{x}{x+5}$, $g(t)=\frac{2}{7-t^2}$
11. $f(x)=\frac{2x}{5-x^2}$, $g(t)=\sqrt{4t+1}$
12. $f(x)=\sqrt{2x+5}$, $g(t)=\frac{10t}{t^2+1}$

In Exercises -, use the given pair of functions to find and simplify expressions for the following functions and state the domain of each using interval notation.

• $(g\circ f)(x)$
• $(f\circ g)(t)$
• $(f\circ f)(x)$
• $f(x)=2x+3$, $g(t)=t^2-9$
• $f(x)=x^2-x+1$, $g(t)=3t-5$
• $f(x)=x^2-4$, $g(t)=|t|$
• $f(x)=3x-5$, $g(t)=\sqrt{t}$
• $f(x)=|x+1|$, $g(t)=\sqrt{t}$
• $f(x)=3-x^2$, $g(t)=\sqrt{t+1}$
• $f(x)=|x|$, $g(t)=\sqrt{4-t}$
• $f(x)=x^2-x-1$, $g(t)=\sqrt{t-5}$
• $f(x)=3x-1$, $g(t)=\frac{1}{t+3}$
• $f(x)=\frac{3x}{x-1}$, $g(t)=\frac{t}{t-3}$
• $f(x)=\frac{x}{2x+1}$, $g(t)=\frac{2t+1}{t}$
• $f(x)=\frac{2x}{x^2-4}$, $g(t)=\sqrt{1-t}$

In Exercises -, use $f(x)=-2x$, $g(t)=\sqrt{t}$ and $h(s)=|s|$ to find and simplify expressions for the following functions and state the domain of each using interval notation.

1. $(h\circ g\circ f)(x)$
2. $(h\circ f\circ g)(t)$
3. $(g\circ f\circ h)(s)$
4. $(g\circ h\circ f)(x)$
5. $(f\circ h\circ g)(t)$
6. $(f\circ g\circ h)(s)$

In Exercises -, let $f$ be the function defined by

$$f=\{(-3,4),(-2,2),(-1,0),(0,1),(1,3),(2,4),(3,-1)\}$$

and let $g$ be the function defined by

$$g=\{(-3,-2),(-2,0),(-1,-4),(0,0),(1,-3),(2,1),(3,2)\}.$$

Find the following, if it exists.

1. $(f\circ g)(3)$
2. $f(g(-1))$
3. $(f\circ f)(0)$
4. $(f\circ g)(-3)$
5. $(g\circ f)(3)$
6. $g(f(-3))$
7. $(g\circ g)(-2)$
8. $(g\circ f)(-2)$
9. $g(f(g(0)))$
10. $f(f(f(-1)))$
11. $f(f(f(f(f(1)))))$
12. $\underset{n\text{ times}}{\underbrace{(g\circ g\circ \mathrm{\cdots }\circ g)}}(0)$
13. Find the domain and range of $f\circ g$ and $g\circ f$.

In Exercises -, use the graphs of $y=f(x)$ and $y=g(x)$ below to find the following if it exists.

1. $(g\circ f)(1)$
2. $(f\circ g)(3)$
3. $(g\circ f)(2)$
4. $(f\circ g)(0)$
5. $(f\circ f)(4)$
6. $(g\circ g)(1)$
7. Find the domain and range of $f\circ g$ and $g\circ f$.

In Exercises -, write the given function as a composition of two or more non-identity functions. (There are several correct answers, so check your answer using function composition.)

1. $p(x)={(2x+3)}^3$
2. $P(x)={\left(x^2-x+1\right)}^5$
3. $h(t)=\sqrt{2t-1}$
4. $H(t)=|7-3t|$
5. $r(s)=\frac{2}{5s+1}$
6. $R(s)=\frac{7}{s^2-1}$
7. $q(z)=\frac{|z|+1}{|z|-1}$
8. $Q(z)=\frac{2z^3+1}{z^3-1}$
9. $v(x)=\frac{2x+1}{3-4x}$
10. $w(x)=\frac{x^2}{x^4+1}$
11. Write the function $F(x)=\sqrt{\frac{x^3+6}{x^3-9}}$ as a composition of three or more non-identity functions.
12. Let $$g(x)=-x,h(x)=x+2,j(x)=3x$$ and $k(x)=x-4$. In what order must these functions be composed with $f(x)=\sqrt{x}$ to create $F(x)=3\sqrt{-x+2}-4$?
13. What linear functions could be used to transform $f(x)=x^3$ into $F(x)=-\frac{1}{2}{(2x-7)}^3+1$? What is the proper order of composition?

14. Let $f(x)=3x+1$ and let

    

    . Find expressions for $(f\circ g)(x)$ and $(g\circ f)(x)$.

15. The volume $V$ of a cube is a function of its side length $x$. Let's assume that $x=t+1$ is also a function of time $t$, where $x$ is measured in inches and $t$ is measured in minutes. Find a formula for $V$ as a function of $t$.

16. Suppose a local vendor charges $\$2$ per hot dog and that the number of hot dogs sold per hour $x$ is given by $x(t)=-4t^2+20t+92$, where $t$ is the number of hours since $10$ AM, $0\le t\le 4$.

    1. Find an expression for the revenue per hour $R$ as a function of $x$.
    2. Find and simplify $\left(R\circ x\right)(t)$. What does this represent?
    3. What is the revenue per hour at noon?
    4. Using Example 5.3.5 as a guide, verify $$\frac{\mathrm{\Delta }[R(x)]}{\mathrm{\Delta }x}\cdot \frac{\mathrm{\Delta }[x(t)]}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }[R(t)]}{\mathrm{\Delta }t}$$.

17. The book in Example 5.3.6 plunges into a lake and generates a circular wave pattern. If the waves are tracked as traveling at a constant $0.5$ meters per second $\left(\frac{\text{m}}{\text{s}}\right)$, use Theorem 5.5 to find the rate at which the area of the disturbance is changing with respect to time as the radius changes from $r=1$ to $r=1.1$ meters (m). Be sure to include units on your answer.

    HINT: Recall the area, $A$, enclosed by a circle of radius $r$ is given by $A=\pi r^2$. Here, $\frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}=0.5\frac{\text{m}}{\text{s}}$.

18. Perfectly fine precalculus textbooks which have no Calculus content are being fed into a shredder at a rate of 2 books per minute in order to make room for precalculus textbooks with Calculus content. The shredder creates a pile of debris which is in the shape of a right circular cone whose height is twice its width.

    1. Assume the volume of the conical pile, $V$, is given by $V=\frac{1}{3}\pi r^{2}h$ where $r$ is the radius of the base of the pile and $h$ is the height of the pile. Given the pile is twice as tall as it is wide, show we can write $V=\frac{4}{3}\pi r^3$.
    2. Assuming a typical precalculus textbook is $0.10$ cubic feet $\left({\text{ft}}^3\right)$, use Theorem 5.5 to find the rate of change of the radius of the pile with respect to time as the radius changes from $2$ to $2.1$ feet. Be sure to include units on your answer.
19. Discuss with your classmates how `real-world' processes such as filling out federal income tax forms or computing your final course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense.

Answers

1. For $f(x)=x^2$ and $g(t)=2t+1$,

   • $(g\circ f)(0)=1$
   • $(f\circ g)(-1)=1$
   • $(f\circ f)(2)=16$
   • $(g\circ f)(-3)=19$
   • $(f\circ g)\left(\frac{1}{2}\right)=4$
   • $(f\circ f)(-2)=16$

2. For $f(x)=4-x$ and $g(t)=1-t^2$,

   • $(g\circ f)(0)=-15$
   • $(f\circ g)(-1)=4$
   • $(f\circ f)(2)=2$
   • $(g\circ f)(-3)=-48$
   • $(f\circ g)\left(\frac{1}{2}\right)=\frac{13}{4}$
   • $(f\circ f)(-2)=-2$

3. For $f(x)=4-3x$ and $g(t)=|t|$,

   • $(g\circ f)(0)=4$
   • $(f\circ g)(-1)=1$
   • $(f\circ f)(2)=10$
   • $(g\circ f)(-3)=13$
   • $(f\circ g)\left(\frac{1}{2}\right)=\frac{5}{2}$
   • $(f\circ f)(-2)=-26$

4. For $f(x)=|x-1|$ and $g(t)=t^2-5$,

   • $(g\circ f)(0)=-4$
   • $(f\circ g)(-1)=5$
   • $(f\circ f)(2)=0$
   • $(g\circ f)(-3)=11$
   • $(f\circ g)\left(\frac{1}{2}\right)=\frac{23}{4}$
   • $(f\circ f)(-2)=2$

5. For $f(x)=4x+5$ and $g(t)=\sqrt{t}$,

   • $(g\circ f)(0)=\sqrt{5}$
   • $(f\circ g)(-1)$ is not real
   • $(f\circ f)(2)=57$
   • $(g\circ f)(-3)$ is not real
   • $$(f\circ g)\left(\frac{1}{2}\right)=5+2\sqrt{2}$$
   • $(f\circ f)(-2)=-7$

6. For $f(x)=\sqrt{3-x}$ and $g(t)=t^2+1$,

   • $(g\circ f)(0)=4$
   • $(f\circ g)(-1)=1$
   • $(f\circ f)(2)=\sqrt{2}$
   • $(g\circ f)(-3)=7$
   • $(f\circ g)\left(\frac{1}{2}\right)=\frac{\sqrt{7}}{2}$
   • $$(f\circ f)(-2)=\sqrt{3-\sqrt{5}}$$

7. For $f(x)=6-x-x^2$ and $g(t)=t\sqrt{t+10}$,

   • $(g\circ f)(0)=24$
   • $(f\circ g)(-1)=0$
   • $(f\circ f)(2)=6$
   • $(g\circ f)(-3)=0$
   • $$(f\circ g)\left(\frac{1}{2}\right)=\frac{27-2\sqrt{42}}{8}$$
   • $(f\circ f)(-2)=-14$

8. For $f(x)=\sqrt[3]{x+1}$ and $g(t)=4t^2-t$,

   • $(g\circ f)(0)=3$
   • $(f\circ g)(-1)=\sqrt[3]{6}$
   • $$(f\circ f)(2)=\sqrt[3]{\sqrt[3]{3}+1}$$
   • $$(g\circ f)(-3)=4\sqrt[3]{4}+\sqrt[3]{2}$$
   • $(f\circ g)\left(\frac{1}{2}\right)=\frac{\sqrt[3]{12}}{2}$
   • $(f\circ f)(-2)=0$

9. For $f(x)=\frac{3}{1-x}$ and $g(t)=\frac{4t}{t^2+1}$,

   • $(g\circ f)(0)=\frac{6}{5}$
   • $(f\circ g)(-1)=1$
   • $(f\circ f)(2)=\frac{3}{4}$
   • $(g\circ f)(-3)=\frac{48}{25}$
   • $(f\circ g)\left(\frac{1}{2}\right)=-5$
   • $(f\circ f)(-2)$ is undefined

10. For $f(x)=\frac{x}{x+5}$ and $g(t)=\frac{2}{7-t^2}$,

    • $(g\circ f)(0)=\frac{2}{7}$
    • $(f\circ g)(-1)=\frac{1}{16}$
    • $(f\circ f)(2)=\frac{2}{37}$
    • $(g\circ f)(-3)=\frac{8}{19}$
    • $(f\circ g)\left(\frac{1}{2}\right)=\frac{8}{143}$
    • $(f\circ f)(-2)=-\frac{2}{13}$

11. For $f(x)=\frac{2x}{5-x^2}$ and $g(t)=\sqrt{4t+1}$,

    • $(g\circ f)(0)=1$
    • $(f\circ g)(-1)$ is not real
    • $(f\circ f)(2)=-\frac{8}{11}$
    • $(g\circ f)(-3)=\sqrt{7}$
    • $(f\circ g)\left(\frac{1}{2}\right)=\sqrt{3}$
    • $(f\circ f)(-2)=\frac{8}{11}$

12. For $f(x)=\sqrt{2x+5}$ and $g(t)=\frac{10t}{t^2+1}$,

    • $(g\circ f)(0)=\frac{5\sqrt{5}}{3}$
    • $(f\circ g)(-1)$ is not real
    • $(f\circ f)(2)=\sqrt{11}$
    • $(g\circ f)(-3)$ is not real
    • $(f\circ g)\left(\frac{1}{2}\right)=\sqrt{13}$
    • $(f\circ f)(-2)=\sqrt{7}$

13. For $f(x)=2x+3$ and $g(t)=t^2-9$

    • $$(g\circ f)(x)=4x^2+12x$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ g)(t)=2t^2-15$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $(f\circ f)(x)=4x+9$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

14. For $f(x)=x^2-x+1$ and $g(t)=3t-5$

    • $$(g\circ f)(x)=3x^2-3x-2$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ g)(t)=9t^2-33t+31$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ f)(x)=x^4-2x^3+2x^2-x+1$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

15. For $f(x)=x^2-4$ and $g(t)=|t|$

    • $$(g\circ f)(x)=|x^2-4|$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ g)(t)={|t|}^2-4=t^2-4$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ f)(x)=x^4-8x^2+12$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

16. For $f(x)=3x-5$ and $g(t)=\sqrt{t}$

    • $$(g\circ f)(x)=\sqrt{3x-5}$$, domain: $[\frac{5}{3},\mathrm{\infty })$
    • $$(f\circ g)(t)=3\sqrt{t}-5$$, domain: $[0,\mathrm{\infty })$
    • $$(f\circ f)(x)=9x-20$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

17. For $f(x)=|x+1|$ and $g(t)=\sqrt{t}$

    • $$(g\circ f)(x)=\sqrt{|x+1|}$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
    • $$(f\circ g)(t)=|\sqrt{t}+1|=\sqrt{t}+1$$, domain: $[0,\mathrm{\infty })$
    • $$(f\circ f)(x)=||x+1|+1|=|x+1|+1$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

18. For $f(x)=3-x^2$ and $g(t)=\sqrt{t+1}$

    • $$(g\circ f)(x)=\sqrt{4-x^2}$$, domain: $[-2,2]$
    • $(f\circ g)(t)=2-t$, domain: $[-1,\mathrm{\infty })$
    • $$(f\circ f)(x)=-x^4+6x^2-6$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

19. For $f(x)=|x|$ and $g(t)=\sqrt{4-t}$

    • $$(g\circ f)(x)=\sqrt{4-|x|}$$, domain: $[-4,4]$
    • $$(f\circ g)(t)=|\sqrt{4-t}|=\sqrt{4-t}$$, domain: $(-\mathrm{\infty },4]$
    • $$(f\circ f)(x)=\Vert x\Vert =|x|$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

20. For $f(x)=x^2-x-1$ and $g(t)=\sqrt{t-5}$

    • $$(g\circ f)(x)=\sqrt{x^2-x-6}$$, domain: $(-\mathrm{\infty },-2]\cup [3,\mathrm{\infty })$
    • $$(f\circ g)(t)=t-6-\sqrt{t-5}$$, domain: $[5,\mathrm{\infty })$
    • $$(f\circ f)(x)=x^4-2x^3-2x^2+3x+1$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

21. For $f(x)=3x-1$ and $g(t)=\frac{1}{t+3}$

    • $(g\circ f)(x)=\frac{1}{3x+2}$, domain: $(-\mathrm{\infty },-\frac{2}{3})\cup (-\frac{2}{3},\mathrm{\infty })$
    • $(f\circ g)(t)=-\frac{t}{t+3}$, domain: $$(-\mathrm{\infty },-3)\cup (-3,\mathrm{\infty })$$
    • $(f\circ f)(x)=9x-4$, domain: $(-\mathrm{\infty },\mathrm{\infty })$

22. For $f(x)=\frac{3x}{x-1}$ and $g(t)=\frac{t}{t-3}$

    • $(g\circ f)(x)=x$, domain: $(-\mathrm{\infty },1)\cup (1,\mathrm{\infty })$
    • $(f\circ g)(t)=t$, domain: $(-\mathrm{\infty },3)\cup (3,\mathrm{\infty })$
    • $(f\circ f)(x)=\frac{9x}{2x+1}$, domain: $$(-\mathrm{\infty },-\frac{1}{2})\cup (-\frac{1}{2},1)\cup (1,\mathrm{\infty })$$

23. For $f(x)=\frac{x}{2x+1}$ and $g(t)=\frac{2t+1}{t}$

    • $(g\circ f)(x)=\frac{4x+1}{x}$, domain: $$(-\mathrm{\infty },-\frac{1}{2})\cup (-\frac{1}{2},0),\cup (0,\mathrm{\infty })$$
    • $(f\circ g)(t)=\frac{2t+1}{5t+2}$, domain: $$(-\mathrm{\infty },-\frac{2}{5})\cup (-\frac{2}{5},0)\cup (0,\mathrm{\infty })$$
    • $(f\circ f)(x)=\frac{x}{4x+1}$, domain: $$(-\mathrm{\infty },-\frac{1}{2})\cup (-\frac{1}{2},-\frac{1}{4})\cup (-\frac{1}{4},\mathrm{\infty })$$

24. For $f(x)=\frac{2x}{x^2-4}$ and $g(t)=\sqrt{1-t}$

    • $$(g\circ f)(x)=\sqrt{\frac{x^2-2x-4}{x^2-4}}$$, domain: $$(-\mathrm{\infty },-2)\cup [1-\sqrt{5},2)\cup [1+\sqrt{5},\mathrm{\infty })$$
    • $$(f\circ g)(t)=-\frac{2\sqrt{1-t}}{t+3}$$, domain: $(-\mathrm{\infty },-3)\cup (-3,1]$
    • $$(f\circ f)(x)=\frac{4x-x^3}{x^4-9x^2+16}$$, domain: $$(-\mathrm{\infty },-\frac{1+\sqrt{17}}{2})\cup (-\frac{1+\sqrt{17}}{2},-2)\cup (-2,\frac{1-\sqrt{17}}{2})\cup (\frac{1-\sqrt{17}}{2},\frac{-1+\sqrt{17}}{2})\cup (\frac{-1+\sqrt{17}}{2},2)\cup (2,\frac{1+\sqrt{17}}{2})\cup (\frac{1+\sqrt{17}}{2},\mathrm{\infty })$$
25. $$(h\circ g\circ f)(x)=|\sqrt{-2x}|=\sqrt{-2x}$$, domain: $(-\mathrm{\infty },0]$
26. $$(h\circ f\circ g)(t)=|-2\sqrt{t}|=2\sqrt{t}$$, domain: $[0,\mathrm{\infty })$
27. $(g\circ f\circ h)(s)=\sqrt{-2|s|}$, domain: $\{0\}$
28. $$(g\circ h\circ f)(x)=\sqrt{|-2x|}=\sqrt{2|x|}$$, domain: $(-\mathrm{\infty },\mathrm{\infty })$
29. $$(f\circ h\circ g)(t)=-2|\sqrt{t}|=-2\sqrt{t}$$, domain: $[0,\mathrm{\infty })$
30. $(f\circ g\circ h)(s)=-2\sqrt{|s|}$,, domain: $(-\mathrm{\infty },\mathrm{\infty })$
31. $$(f\circ g)(3)=f(g(3))=f(2)=4$$
32. $f(g(-1))=f(-4)$ which is undefined
33. $$(f\circ f)(0)=f(f(0))=f(1)=3$$
34. $$(f\circ g)(-3)=f(g(-3))=f(-2)=2$$
35. $$(g\circ f)(3)=g(f(3))=g(-1)=-4$$
36. $g(f(-3))=g(4)$ which is undefined
37. $$(g\circ g)(-2)=g(g(-2))=g(0)=0$$
38. $$(g\circ f)(-2)=g(f(-2))=g(2)=1$$
39. $$g(f(g(0)))=g(f(0))=g(1)=-3$$
40. $$f(f(f(-1)))=f(f(0))=f(1)=3$$
41. $$f(f(f(f(f(1)))))=f(f(f(f(3))))=f(f(f(-1)))=f(f(0))=f(1)=3$$
42. $\underset{n\text{ times}}{\underbrace{(g\circ g\circ \mathrm{\cdots }\circ g)}}(0)=0$
43. • The domain of $f\circ g$ is $\{-3,-2,0,1,2,3\}$ and the range of $f\circ g$ is $\{1,2,3,4\}$.
    • The domain of $g\circ f$ is $\{-2,-1,0,1,3\}$ and the range of $g\circ f$ is $\{-4,-3,0,1,2\}$.
44. $(g\circ f)(1)=3$
45. $(f\circ g)(3)=1$
46. $(g\circ f)(2)=0$
47. $(f\circ g)(0)=1$
48. $(f\circ f)(4)=1$
49. $(g\circ g)(1)=0$
50. • The domain of $f\circ g$ is $[0,3]$ and the range of $f\circ g$ is $[1,4.5]$.
    • The domain of $g\circ f$ is $[0,2]\cup [3,4]$ and the range is $[0,3]$.
51. Let $f(x)=2x+3$ and $g(x)=x^3$, then $p(x)=(g\circ f)(x)$.
52. Let $f(x)=x^2-x+1$ and $g(x)=x^5$, $P(x)=(g\circ f)(x)$.
53. Let $f(t)=2t-1$ and $g(t)=\sqrt{t}$, then $h(t)=(g\circ f)(t)$.
54. Let $f(t)=7-3t$ and $g(t)=|t|$, then $H(t)=(g\circ f)(t)$.
55. Let $f(s)=5s+1$ and $g(s)=\frac{2}{s}$, then $r(s)=(g\circ f)(s)$.
56. Let $f(s)=s^2-1$ and $g(s)=\frac{7}{s}$, then $R(s)=(g\circ f)(s)$.
57. Let $f(z)=|z|$ and $g(z)=\frac{z+1}{z-1}$, then $q(z)=(g\circ f)(z)$.
58. Let $f(z)=z^3$ and $g(z)=\frac{2z+1}{z-1}$, then $Q(z)=(g\circ f)(z)$.
59. Let $f(x)=2x$ and $g(x)=\frac{x+1}{3-2x}$, then $v(x)=(g\circ f)(x)$.
60. Let $f(x)=x^2$ and $g(x)=\frac{x}{x^2+1}$, then $w(x)=(g\circ f)(x)$.
61. $$F(x)=\sqrt{\frac{x^3+6}{x^3-9}}=(h(g(f(x)))$$ where $f(x)=x^3,g(x)=\frac{x+6}{x-9}$ and $h(x)=\sqrt{x}$.
62. $$F(x)=3\sqrt{-x+2}-4=k(j(f(h(g(x)))))$$
63. One solution is $$F(x)=-\frac{1}{2}{(2x-7)}^3+1=k(j(f(h(g(x)))))$$ where $$g(x)=2x,h(x)=x-7,j(x)=-\frac{1}{2}x$$ and $k(x)=x+1$. You could also have $F(x)=H(f(G(x)))$ where $G(x)=2x-7$ and $H(x)=-\frac{1}{2}x+1$.

64. 

    and

    

65. $V(x)=x^3$ so $V(x(t))={(t+1)}^3$
66. 1. $R(x)=2x$
    2. $$\left(R\circ x\right)(t)=-8t^2+40t+184$$, $0\le t\le 4$. This gives the revenue per hour as a function of time.
    3. Noon corresponds to $t=2$, so $\left(R\circ x\right)(2)=232$. The hourly revenue at noon is $\$232$ per hour.

    4. $\frac{\mathrm{\Delta }[R(x)]}{\mathrm{\Delta }x}=2$, $$\frac{\mathrm{\Delta }[x(t)]}{\mathrm{\Delta }t}=-8t-4\mathrm{\Delta }t+20$$.

       $$\frac{\mathrm{\Delta }[R(x)]}{\mathrm{\Delta }x}\cdot \frac{\mathrm{\Delta }[x(t)]}{\mathrm{\Delta }t}=(2)(-8t+4\mathrm{\Delta }t+20)=-16t+8\mathrm{\Delta }t+40=\frac{\mathrm{\Delta }[R(t)]}{\mathrm{\Delta }t}\mathrm{✓}$$

67. $\frac{\mathrm{\Delta }A}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }A}{\mathrm{\Delta }r}\cdot \frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}$. $$\frac{\mathrm{\Delta }A}{\mathrm{\Delta }r}=\frac{A(1.1)-A(1)}{1.1-1}=\frac{\pi {(1.1)}^2-\pi {(1)}^2}{0.1}=2.1\frac{{\text{m}}^2}{\text{m}}$$, $\frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}=0.5\frac{\text{m}}{\text{s}}$.

    Hence, $$\frac{\mathrm{\Delta }A}{\mathrm{\Delta }t}=\left(2.1\frac{{\text{m}}^2}{\text{m}}\right)\left(0.5\frac{\text{m}}{\text{s}}\right)=1.05\frac{{\text{m}}^2}{\text{s}}$$

68. 1. The `width' of the pile is the diameter of the circular base of the pile. Since the diameter of a circle is twice the radius, $h=2(2r)=4r$. Hence, $$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2(4r)=\frac{4}{3}\pi r^3$$.

    2. We have $\frac{\mathrm{\Delta }V}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }V}{\mathrm{\Delta }r}\cdot \frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}$. Two textbooks per minute into the shredder amounts to the volume of the cone increasing at a rate of $2(0.1)=0.2\frac{{\text{ft}}^3}{\text{min}}$. $$\frac{\mathrm{\Delta }V}{\mathrm{\Delta }r}=\frac{V(2.1)-V(2)}{2.1-2}=16.81\overline{3}\pi \frac{{\text{ft}}^3}{\text{ft}}$$.

       Hence, $0.2\frac{{\text{ft}}^3}{\text{min}}=16.81\overline{3}\pi \frac{{\text{ft}}^3}{\text{ft}}\frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}$ so $$\frac{\mathrm{\Delta }r}{\mathrm{\Delta }t}=\frac{0.2}{16.81\overline{3}\pi }\approx 0.0038\frac{\text{ft}}{\text{min}}$$.