Function composition and decomposition

Learning goals

Evaluates compositions of functions.
Determines domains of composite functions.
Decomposes a function into meaningful component functions when possible.
Interprets composition in context or as successive transformations.

Difficulty: 1 – 5

Click problems to add them to your worksheet.

Function composition and decomposition 1

In the composition $(f\circ g)(4)$, which function is applied to $4$ first: $f$ or $g$?

Hints

- Rewrite the composition notation using nested function notation. - The inner function acts on the original input first.

Solution

1. Composition notation means $(f\circ g)(4)=f(g(4))$. 2. Therefore, $g$ is applied first.

Answer

$g$ is applied first.

Function composition and decomposition 2

Suppose $f(3)=8$ and $g(8)=1$. Find $g(f(3))$.

Hints

- Start with the inner function value that is already given. - Feed that output into the outer function.

Solution

1. The inner value is given directly: $f(3)=8$. 2. Use that output as the input of $g$: $g(8)=1$. 3. Therefore, $g(f(3))=1$.

Answer

$1$

Function composition and decomposition 3

Let $f(x)=x+3$ and $g(x)=x^2$. Starting with $2$, first apply $f$, then apply $g$. Find $g(f(2))$.

Hints

- Which function gives the output you must use first? - Use the result of the inner function as the input of the outer function.

Solution

1. The first function gives $f(2)=2+3=5$. 2. Use that output in the second function: $g(f(2))=g(5)=25$.

Answer

$25$

Function composition and decomposition 4

Let $f(x)=\sqrt{x}$ and $g(x)=x+7$. Which rule gives $f(g(x))$: $\sqrt{x+7}$ or $\sqrt{x}+7$? Explain which function acts first.

Hints

- In $f(g(x))$, which function receives the original input? - What does the second function do to the first one's output?

Solution

1. Apply $g$ to the input first, giving $g(x)=x+7$. 2. Apply $f$ to that result: $f(g(x))=\sqrt{x+7}$.

Answer

$f(g(x))=\sqrt{x+7}$. Apply $g$ first, then $f$.

Function composition and decomposition 5

Let $f(x)=\sqrt{x}$ and $g(x)=x-3$. What values of $x$ are allowed in $f(g(x))$?

Hints

- Write the inner function's output inside the outer function. - What must be true of a real square root's input?

Solution

1. The composition is $\sqrt{x-3}$. 2. A real square root requires $x-3\ge0$, so $x\ge3$.

Answer

$x\ge3$

Function composition and decomposition 6

Let $h(x)=3x^2+1$ and $f(u)=3u+1$. Find a function $g$ so that $h(x)=f(g(x))$.

Hints

- Compare the form of $h$ with what the outer rule does to an input. - What expression must be fed into $f$ to produce the squared term?

Solution

1. The outer function multiplies its input by $3$ and adds $1$. 2. For the result to be $3x^2+1$, its input must be $x^2$. Thus $g(x)=x^2$.

Answer

$g(x)=x^2$

Function composition and decomposition 7

Let $u(x)=\sqrt{x+2}$ and $v(x)=e^x$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- Substitute the entire inner function into the outer function. - Check both the domain of the inner function and the input restrictions of the outer function. - Recall the domains of exponential and square-root functions.

Solution

1. For $f=u\circ v$, $f(x)=u(v(x))=\sqrt{e^x+2}$. Since $e^x>0$ for every real $x$, the radicand is always positive. Therefore, $D_f=\mathbb{R}$. 2. For $g=v\circ u$, $g(x)=v(u(x))=e^{\sqrt{x+2}}$. The square root requires $x+2\ge0$, so $D_g=[-2, \infty)$.

Answer

$f(x)=\sqrt{e^x+2}$, $D_f=\mathbb{R}$ $g(x)=e^{\sqrt{x+2}}$, $D_g=[-2, \infty)$

Function composition and decomposition 8

Let $f(x)=3x^2-4$ and $g(x)=2-x$. Evaluate each expression. a) $f(g(3))$ b) $g(f(-2))$ c) $(f\circ g)(1)$ d) $(g\circ g)(5)$

Hints

- Evaluate the inner function first. - Use the result as the input of the outer function. - The notation $f\circ g$ means $f(g(x))$.

Solution

1. $g(3)=-1$, so $f(g(3))=f(-1)=3\cdot(-1)^2-4=-1$. 2. $f(-2)=3\cdot(-2)^2-4=8$, so $g(f(-2))=g(8)=-6$. 3. $g(1)=1$, so $(f\circ g)(1)=f(1)=-1$. 4. $g(5)=-3$, so $(g\circ g)(5)=g(-3)=5$.

Answer

a) $-1$ b) $-6$ c) $-1$ d) $5$

Function composition and decomposition 9

Let $p(x)=e^{x-1}$ and $q(x)=4x+1$. Evaluate each expression. a) $p(q(0))$ b) $q(p(1))$ c) $(p\circ q)(0.25)$ d) $q(q(-1))$

Hints

- Evaluate the inner function first. - Recall that $e^0=1$. - Pay attention to the order in which the functions are applied.

Solution

1. $q(0)=1$, so $p(q(0))=p(1)=e^0=1$. 2. $p(1)=1$, so $q(p(1))=q(1)=5$. 3. $q(0.25)=2$, so $(p\circ q)(0.25)=p(2)=e$. 4. $q(-1)=-3$, so $q(q(-1))=q(-3)=-11$.

Answer

a) $1$ b) $5$ c) $e$ d) $-11$

Function composition and decomposition 10

Let $f(x)=e^x$ and $g(x)=x^2+2$. a) Find equations for $u=f\circ g$ and $v=g\circ f$. b) Show that function composition is not commutative by comparing $u(0)$ and $v(0)$.

Hints

- Identify the inner and outer function in each composition. - One input where the outputs differ is enough to prove that two functions are not identical.

Solution

1. Composing in the two orders gives $u(x)=f(g(x))=e^{x^2+2}$ and $v(x)=g(f(x))=(e^x)^2+2=e^{2x}+2$. 2. At $x=0$, $u(0)=e^2$ and $v(0)=e^0+2=3$. 3. Since $e^2\ne3$, the functions are not equal. Therefore, $f\circ g\ne g\circ f$.

Answer

a) $u(x)=e^{x^2+2}$ and $v(x)=e^{2x}+2$ b) $u(0)=e^2$ and $v(0)=3$, so $f\circ g\ne g\circ f$.

Function composition and decomposition 11

Let $u(x)=3x-2$, $v(x)=x^2$, and $w(x)=x+1$. a) Find formulas for $f=u\circ v\circ w$ and $g=w\circ v\circ u$. b) Evaluate $f(2)$.

Hints

- In a composition, which function acts on the original input first? - Keep the full inner expression inside the next function. - Does changing the order of the same three functions change the result?

Solution

1. Work from the inside out: $f(x)=u(v(x+1))=u((x+1)^2)=3(x+1)^2-2$. 2. Reversing the order gives $g(x)=w(v(3x-2))=w((3x-2)^2)=(3x-2)^2+1$. 3. Substitute $x=2$: $f(2)=3\cdot3^2-2=25$.

Answer

a) $f(x)=3(x+1)^2-2$; $g(x)=(3x-2)^2+1$ b) $f(2)=25$

Function composition and decomposition 12

For each function, find an inner function $v$ and an outer function $u$ such that $f=u\circ v$. a) $f(x)=e^{3x^2-5}$ b) $f(x)=\frac{2}{|x|-4}$ c) $f(x)=\ln(\frac{2x+1}{x^2})$ d) $f(x)=4(\sqrt{x}+1)^5$

Hints

- Identify the first operation performed on the input. - Treat the inner expression as a single value to see the outside operation. - A different split may also work; check yours by recomposing.

Solution

1. For a), choose $v(x)=3x^2-5$ and $u(x)=e^x$. 2. For b), choose $v(x)=|x|-4$ and $u(x)=\frac{2}{x}$. 3. For c), choose $v(x)=\frac{2x+1}{x^2}$ and $u(x)=\ln(x)$. 4. For d), choose $v(x)=\sqrt{x}+1$ and $u(x)=4x^5$. 5. In each case, inserting $v(x)$ into $u$ reproduces the given function on its domain.

Answer

a) $v(x)=3x^2-5$, $u(x)=e^x$ b) $v(x)=|x|-4$, $u(x)=\frac{2}{x}$ c) $v(x)=\frac{2x+1}{x^2}$, $u(x)=\ln(x)$ d) $v(x)=\sqrt{x}+1$, $u(x)=4x^5$

Function composition and decomposition 13

Consider $h(x)=\sqrt{e^{x^2}+2}$. a) Find functions $f$, $g$, and $k$ such that $h=f\circ g\circ k$. b) Give a different decomposition $h=u\circ v$ with a square-root outer function.

Hints

- Which operation acts on $x$ first? - Think of the expression immediately inside the root as one output. - Check by substituting from the inside function to the outside function.

Solution

1. Choose $k(x)=x^2$, $g(x)=e^x+2$, and $f(x)=\sqrt{x}$. Then $f(g(k(x)))=\sqrt{e^{x^2}+2}$. 2. For two functions, choose $v(x)=e^{x^2}+2$ and $u(x)=\sqrt{x}$, giving $h=u\circ v$.

Answer

a) One choice: $k(x)=x^2$, $g(x)=e^x+2$, $f(x)=\sqrt{x}$ b) $v(x)=e^{x^2}+2$, $u(x)=\sqrt{x}$

Function composition and decomposition 14

For each function, find an inner function $v$ and an outer function $u$ so that $f=u\circ v$. a) $f(x)=\ln(3x-4)$ b) $f(x)=\sqrt{5-x^2}$ c) $f(x)=(e^x+2)^5$ d) $f(x)=e^{\frac{1}{x}}$

Hints

- Identify the operation performed on the original input first. - Which operation turns that intermediate output into the final result? - Check each choice by writing $u(v(x))$.

Solution

1. For a), choose $v(x)=3x-4$, $u(x)=\ln(x)$. 2. For b), choose $v(x)=5-x^2$, $u(x)=\sqrt{x}$. 3. For c), choose $v(x)=e^x+2$, $u(x)=x^5$. 4. For d), choose $v(x)=\frac{1}{x}$, $u(x)=e^x$.

Answer

a) $v(x)=3x-4$, $u(x)=\ln(x)$ b) $v(x)=5-x^2$, $u(x)=\sqrt{x}$ c) $v(x)=e^x+2$, $u(x)=x^5$ d) $v(x)=\frac{1}{x}$, $u(x)=e^x$

Function composition and decomposition 15

For each function $f$, find functions $u$, $v$, and $w$ such that $f(x)=u(v(w(x)))$. a) $f(x)=\frac{1}{(5x+2)^3}$ b) $f(x)=\sqrt{|x^2-4|}$ c) $f(x)=(e^{x-3})^4$

Hints

- Identify the operation acting directly on the original input. - Look for a second operation that transforms that result before the final operation. - Substitute your chosen functions one inside another to check.

Solution

1. For a), choose $w(x)=5x+2$, $v(x)=x^3$, and $u(x)=\frac{1}{x}$. 2. For b), choose $w(x)=x^2-4$, $v(x)=|x|$, and $u(x)=\sqrt{x}$. 3. For c), choose $w(x)=x-3$, $v(x)=e^x$, and $u(x)=x^4$. 4. Substituting each inner output into the next function gives the stated formula.

Answer

a) $w(x)=5x+2$, $v(x)=x^3$, $u(x)=\frac{1}{x}$ b) $w(x)=x^2-4$, $v(x)=|x|$, $u(x)=\sqrt{x}$ c) $w(x)=x-3$, $v(x)=e^x$, $u(x)=x^4$

Function composition and decomposition 16

For each function $f$, find functions $u$, $v$, and $w$ such that $f=u\circ v\circ w$. a) $f(x)=\sqrt[3]{2^{2x}}$ b) $f(x)=\frac{1}{\sqrt{x^2+4x}}$ c) $f(x)=(\frac{1}{x^2+1}+1)^5$

Hints

- Try naming the first operation applied to $x$. - What operation comes between that first step and the final step? - Does composing your choices in the order $w$, then $v$, then $u$ recover the original function?

Solution

1. For a), choose $w(x)=2x$, $v(x)=2^x$, and $u(x)=\sqrt[3]{x}$. 2. For b), choose $w(x)=x^2+4x$, $v(x)=\sqrt{x}$, and $u(x)=\frac{1}{x}$. 3. For c), choose $w(x)=x^2+1$, $v(x)=\frac{1}{x}$, and $u(x)=(x+1)^5$.

Answer

a) $w(x)=2x$, $v(x)=2^x$, $u(x)=\sqrt[3]{x}$ b) $w(x)=x^2+4x$, $v(x)=\sqrt{x}$, $u(x)=\frac{1}{x}$ c) $w(x)=x^2+1$, $v(x)=\frac{1}{x}$, $u(x)=(x+1)^5$

Function composition and decomposition 17

Decompose $f(x)=\left(x+\frac{1}{x}\right)^3$, where $x\ne0$, into an inner function $v$ and an outer function $u$ such that $f=u\circ v$.

Hints

- Identify the expression evaluated first. - The final operation usually defines the outer function. - Check by composing the two functions.

Solution

1. The expression inside the parentheses is evaluated first, so choose $v(x)=x+\frac{1}{x}$. 2. The final operation is cubing, so choose $u(x)=x^3$. 3. Then $u(v(x))=\left(x+\frac{1}{x}\right)^3=f(x)$.

Answer

One possible answer is $v(x)=x+\frac{1}{x}$ and $u(x)=x^3$.

Function composition and decomposition 18

Consider $f(x)=\sqrt{1+(\ln(x+1))^2}$, with $x>-1$. Give a decomposition $f=u\circ v$.

Hints

- There may be more than one valid decomposition. - What expression could you treat as one intermediate output? - What operation would turn that intermediate value into the final result?

Solution

1. Choose the inner function $v(x)=\ln(x+1)$, which is defined for $x>-1$. 2. Choose the outer function $u(t)=\sqrt{1+t^2}$. 3. Then $u(v(x))=\sqrt{1+(\ln(x+1))^2}=f(x)$.

Answer

One choice is $v(x)=\ln(x+1)$ and $u(x)=\sqrt{1+x^2}$.

Function composition and decomposition 19

The figure shows the graphs of functions $f$ and $g$. Define $h=f\circ g$ and $k=g\circ f$. Use the graphs to evaluate each expression. a) $h(0)$ b) $h(2)$ c) $h(4)$ d) $k(-2)$ e) $k(0)$ f) $k(2)$

Hints

- Evaluate the inner function first. - Use the inner output as the x-coordinate for the outer graph. - Pay attention to the order of composition.

Solution

1. For $h(x)=f(g(x))$, read $g(x)$ first and then use that result as the input of $f$. The graph gives $g(0)=2$, $g(2)=0$, and $g(4)=-2$. Therefore, $h(0)=f(2)=0$, $h(2)=f(0)=-2$, and $h(4)=f(-2)=0$. 2. For $k(x)=g(f(x))$, read $f(x)$ first. The graph gives $f(-2)=0$, $f(0)=-2$, and $f(2)=0$. Therefore, $k(-2)=g(0)=2$, $k(0)=g(-2)=4$, and $k(2)=g(0)=2$.

Answer

a) $0$ b) $-2$ c) $0$ d) $2$ e) $4$ f) $2$

Function composition and decomposition 20

The figure shows the graphs of $f$ and $g$. Compare the two compositions at $x=2$: 1) $h_1(2)=f(g(2))$ 2) $h_2(2)=g(f(2))$ Determine whether $h_1(2)=h_2(2)$.

Hints

- Evaluate the inner function first. - Use the first output as the x-value on the second graph. - Composition is generally order-dependent.

Solution

1. From the graph, $g(2)=0$ and $f(0)=3$. Therefore, $h_1(2)=f(g(2))=3$. 2. From the graph, $f(2)=1$ and $g(1)=-0.5$. Therefore, $h_2(2)=g(f(2))=-0.5$. 3. Since $3\ne-0.5$, the two compositions are not equal at $x=2$.

Answer

1) $h_1(2)=3$ 2) $h_2(2)=-0.5$ $h_1(2)\ne h_2(2)$

Function composition and decomposition 21

The figure shows the graph of a positive function $h$. Define $f(x)=\ln(h(x))$. Use the graph to identify the x-value $x_0$ where $h$ has a local maximum. Then explain without calculation why $f$ has a local maximum at the same x-value.

Hints

- Locate the peak of the displayed graph. - Recall how $\ln(u)$ changes as a positive input $u$ increases. - A strictly increasing outer function preserves comparisons between nearby function values.

Solution

1. The graph shows a local maximum of $h$ at $x_0=4$. 2. The natural logarithm is strictly increasing on its domain. Therefore, it preserves the order of positive function values: if $h(x_1)<h(x_2)$, then $\ln(h(x_1))<\ln(h(x_2))$. 3. Near $x=4$, the values of $h$ are no greater than $h(4)$. Applying the increasing logarithm preserves that comparison, so nearby values of $f$ are no greater than $f(4)$. Thus $f$ also has a local maximum at $x_0=4$.

Answer

$x_0=4$. Because $\ln(u)$ is strictly increasing for $u>0$, it preserves the location and type of the local maximum.

Function composition and decomposition 22

Function $c$ accepts an input measured in minutes and returns an output measured in dollars. Function $d$ accepts an input measured in dollars and returns an output measured in reward points. Starting with a ride time $t$ measured in minutes, which composition is meaningful: $d(c(t))$ or $c(d(t))$? Explain by tracking the units through the two functions.

Hints

- Track the unit entering and leaving each function. - Ask which rule can accept the original ride-time input. - The inner function must produce the kind of quantity that the outer function accepts.

Solution

1. The starting quantity is in minutes, so it can be used as an input to $c$. 2. The output $c(t)$ is in dollars, which is exactly the input unit required by $d$. 3. Therefore, $d(c(t))$ is meaningful. The expression $c(d(t))$ is not meaningful from the stated units because $d$ cannot take the original minute input.

Answer

$d(c(t))$. Minutes enter $c$, dollars come out of $c$, and those dollars can then enter $d$.

Function composition and decomposition 23

A robotics club uses two tables. The first function $p$ gives the number of completed test runs after $h$ hours of practice. The second function $s$ converts a number of completed test runs into a score in points. <table> <tr><th>$h$ (hours)</th><th>$p(h)$ (test runs)</th></tr> <tr><td>2</td><td>18</td></tr> <tr><td>4</td><td>35</td></tr> <tr><td>6</td><td>51</td></tr> </table> <table> <tr><th>$n$ (test runs)</th><th>$s(n)$ (points)</th></tr> <tr><td>51</td><td>32</td></tr> <tr><td>18</td><td>9</td></tr> <tr><td>44</td><td>26</td></tr> <tr><td>35</td><td>20</td></tr> </table> Find $s(p(4))$ and interpret the result, including its unit.

Hints

- Start with the innermost function and use the first table. - The output from the first lookup becomes the input for the second lookup. - Check the units as the quantity moves from one table to the next.

Solution

1. From the first table, $p(4)=35$ test runs. 2. Use that output as the input of $s$. From the second table, $s(35)=20$ points. 3. Therefore, $s(p(4))=20$, meaning that $4$ hours of practice corresponds to a score of $20$ points in this two-stage model.

Answer

$s(p(4))=20$ points. In the model, $4$ hours of practice leads to a score of $20$ points.

Function composition and decomposition 24

Let $f(x)=x+1$ and $g(x)=2x$. Find $f(g(2))$ and $g(f(2))$. What do the results show about composition order?

Hints

- For $f(g(2))$, which function must act on $2$ first? - Work out both orders separately before comparing them.

Solution

1. Apply $g$ first: $g(2)=4$, so $f(g(2))=f(4)=5$. 2. Apply $f$ first: $f(2)=3$, so $g(f(2))=g(3)=6$. 3. The two outputs differ, so composition order can affect the result.

Answer

$f(g(2))=5$, while $g(f(2))=6$. Composition order can change the output.

Function composition and decomposition 25

The graphs of functions $f$ and $g$ are shown. Use the graphs to find $f(g(2))$ and $g(f(2))$. Are the outputs equal?

Hints

- Read the value of the inner function at $2$ first. - Use that output as the x-coordinate on the other graph. - Repeat the process in the other order before comparing.

Solution

1. The line gives $g(2)=0$, and the parabola gives $f(0)=0$, so $f(g(2))=0$. 2. The parabola gives $f(2)=4$, and the line gives $g(4)=2$, so $g(f(2))=2$. 3. The outputs are different.

Answer

$f(g(2))=0$ and $g(f(2))=2$; no, they are not equal.

Function composition and decomposition 26

Maya walks $3$ miles per hour in a school walkathon. Her sponsors have pledged $\$5$ per mile, plus an initial $\$20$ donation. Let $d(t)=3t$ be the miles walked in $t$ hours and $R(d)=20+5d$ the dollars raised after $d$ miles. Find $R(d(2))$ and explain what it means.

Hints

- Which quantity does the time rule calculate? - Use the distance as the input for the donation rule. - Does the final result represent miles, hours, or dollars?

Solution

1. In $2$ hours, Maya walks $d(2)=3\cdot2=6$ miles. 2. Use the distance in the donation rule: $R(d(2))=R(6)=20+5\cdot6=50$ dollars. 3. Her sponsors have pledged a total of $\$50$ for the walk.

Answer

$R(d(2))=\$50$, the donations earned after $2$ hours of walking.

Function composition and decomposition 27

Let $f(x)=x^2-1$ and $g(x)=\frac{2}{x+3}$. Find each function and its maximal real domain. a) $f+g$ b) $g\circ f$ c) $f\circ g$

Hints

- Identify the domain of each original function first. - For a sum, both functions must be defined. - For a composition, determine which function is applied first. - A rational expression is undefined when its denominator is zero.

Solution

1. The domains of the original functions are $D_f=\mathbb{R}$ and $D_g=\mathbb{R}\setminus\{-3\}$. 2. For the sum, $(f+g)(x)=x^2-1+\frac{2}{x+3}$. Both functions must be defined, so the domain is $\mathbb{R}\setminus\{-3\}$. 3. For $g\circ f$, $(g\circ f)(x)=\frac{2}{(x^2-1)+3}=\frac{2}{x^2+2}$. The denominator is positive for every real $x$, so the domain is $\mathbb{R}$. 4. For $f\circ g$, $(f\circ g)(x)=\left(\frac{2}{x+3}\right)^2-1=\frac{4}{(x+3)^2}-1$. The only restriction is $x\ne-3$.

Answer

a) $(f+g)(x)=x^2-1+\frac{2}{x+3}$, $D=\mathbb{R}\setminus\{-3\}$ b) $(g\circ f)(x)=\frac{2}{x^2+2}$, $D=\mathbb{R}$ c) $(f\circ g)(x)=\frac{4}{(x+3)^2}-1$, $D=\mathbb{R}\setminus\{-3\}$

Function composition and decomposition 28

Let $u(x)=\ln(x)$ and $v(x)=x^2-4$. Find each function and its maximal real domain. a) $uv$ b) $u\circ v$ c) $v\circ u$

Hints

- The input of a natural logarithm must be positive. - Pay attention to the order of the functions in each composition. - Solve the inequality that makes the logarithm input positive. - Squaring the output of a logarithm adds no new domain restriction.

Solution

1. The original domains are $D_u=(0, \infty)$ and $D_v=\mathbb{R}$. 2. For the product, $(uv)(x)=(x^2-4)\ln(x)$. Both factors must be defined, so the domain is $(0, \infty)$. 3. For $u\circ v$, $(u\circ v)(x)=\ln(x^2-4)$. The logarithm requires $x^2-4>0$, so the domain is $(-\infty, -2)\cup(2, \infty)$. 4. For $v\circ u$, $(v\circ u)(x)=(\ln(x))^2-4$. The logarithm requires $x>0$, and there are no further restrictions, so the domain is $(0, \infty)$.

Answer

a) $(uv)(x)=(x^2-4)\ln(x)$, $D=(0, \infty)$ b) $(u\circ v)(x)=\ln(x^2-4)$, $D=(-\infty, -2)\cup(2, \infty)$ c) $(v\circ u)(x)=(\ln(x))^2-4$, $D=(0, \infty)$

Function composition and decomposition 29

For each pair of functions, find $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition. a) $u(x)=\frac{2}{x-4}$ and $v(x)=x^2$ b) $u(x)=x-5$ and $v(x)=\sqrt{x}$

Hints

- Determine the domain of the inner function first. - The output of the inner function must be in the domain of the outer function. - A denominator cannot be zero. - A square-root radicand must be nonnegative.

Solution

1. For part a, $f(x)=u(v(x))=\frac{2}{x^2-4}$. The denominator cannot equal zero, so $x\ne-2$ and $x\ne2$. Thus $D_f=\mathbb{R}\setminus\{-2, 2\}$. 2. Also, $g(x)=v(u(x))=\left(\frac{2}{x-4}\right)^2=\frac{4}{(x-4)^2}$. The inner function is undefined at $x=4$, so $D_g=\mathbb{R}\setminus\{4\}$. 3. For part b, $f(x)=u(v(x))=\sqrt{x}-5$. The square root requires $x\ge0$, so $D_f=[0, \infty)$. 4. Also, $g(x)=v(u(x))=\sqrt{x-5}$. The radicand must be nonnegative, so $x\ge5$ and $D_g=[5, \infty)$.

Answer

a) $f(x)=\frac{2}{x^2-4}$, $D_f=\mathbb{R}\setminus\{-2, 2\}$; $g(x)=\frac{4}{(x-4)^2}$, $D_g=\mathbb{R}\setminus\{4\}$ b) $f(x)=\sqrt{x}-5$, $D_f=[0, \infty)$; $g(x)=\sqrt{x-5}$, $D_g=[5, \infty)$

Function composition and decomposition 30

Let $f(x)=2x+5$ and $g(x)=3x+c$, where $c\in\mathbb{R}$. a) Find equations for $h=f\circ g$ and $k=g\circ f$. b) Find the value of $c$ for which $h(x)=k(x)$ for every real $x$. c) Choose any other value of $c$ and show that the compositions are not equal.

Hints

- Substitute the entire inner function into the outer function. - Two linear functions are identical when their corresponding coefficients are equal. - Use a simple value of $c$ for the counterexample.

Solution

1. The compositions are $h(x)=f(g(x))=2(3x+c)+5=6x+2c+5$ and $k(x)=g(f(x))=3(2x+5)+c=6x+15+c$. 2. The x-coefficients already agree. Equating the constant terms gives $2c+5=15+c$, so $c=10$. 3. For example, if $c=0$, then $h(x)=6x+5$ and $k(x)=6x+15$. These functions are not equal, so the compositions are not commutative for this value. The same conclusion holds for every $c\ne10$.

Answer

a) $h(x)=6x+2c+5$ and $k(x)=6x+15+c$ b) $c=10$ c) For example, when $c=0$, $h(x)=6x+5\ne6x+15=k(x)$.

Function composition and decomposition 31

Let $u(x)=\ln(x)$ and $v(x)=\frac{1}{x^2-4}$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- Check the domain of the inner function first. - The input of a logarithm must be positive. - A rational expression is undefined when its denominator is zero.

Solution

1. For $f=u\circ v$, $f(x)=\ln\left(\frac{1}{x^2-4}\right)$. The logarithm input must be positive, so $\frac{1}{x^2-4}>0$. This requires $x^2-4>0$, or $|x|>2$. Thus $D_f=(-\infty, -2)\cup(2, \infty)$. 2. For $g=v\circ u$, $g(x)=\frac{1}{(\ln(x))^2-4}$. First, $x>0$. Also, the denominator cannot be zero, so $(\ln(x))^2\ne4$. Therefore, $x\ne e^{-2}$ and $x\ne e^2$. 3. Hence $D_g=(0, e^{-2})\cup(e^{-2}, e^2)\cup(e^2, \infty)$.

Answer

$f(x)=\ln\left(\frac{1}{x^2-4}\right)$, $D_f=(-\infty, -2)\cup(2, \infty)$ $g(x)=\frac{1}{(\ln(x))^2-4}$, $D_g=(0, e^{-2})\cup(e^{-2}, e^2)\cup(e^2, \infty)$

Function composition and decomposition 32

Let $u(x)=\sqrt{x}$ and $v(x)=\frac{x+3}{x-1}$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- A square-root radicand must be nonnegative. - A denominator cannot be zero. - Use a sign chart for the rational inequality. - Begin with the domain of the inner function.

Solution

1. For $f=u\circ v$, $f(x)=\sqrt{\frac{x+3}{x-1}}$. The rational expression must be defined and nonnegative. A sign analysis at the critical values $-3$ and $1$ gives $D_f=(-\infty, -3]\cup(1, \infty)$. 2. For $g=v\circ u$, $g(x)=\frac{\sqrt{x}+3}{\sqrt{x}-1}$. The square root requires $x\ge0$, and the denominator requires $\sqrt{x}\ne1$, so $x\ne1$. 3. Therefore, $D_g=[0, 1)\cup(1, \infty)$.

Answer

$f(x)=\sqrt{\frac{x+3}{x-1}}$, $D_f=(-\infty, -3]\cup(1, \infty)$ $g(x)=\frac{\sqrt{x}+3}{\sqrt{x}-1}$, $D_g=[0, 1)\cup(1, \infty)$

Function composition and decomposition 33

Consider the function $k(x)=\frac{4}{\ln(x^2+e)}$. a) Write $k$ as a composition of three functions in the form $k=f\circ g\circ h$. b) Find the range of $k$.

Hints

- Follow the operations from the inside outward. - Find the smallest possible value of the logarithm input. - Determine how the reciprocal changes as its positive denominator increases.

Solution

1. One decomposition follows the order of operations: $h(x)=x^2+e$, $g(x)=\ln(x)$, and $f(x)=\frac{4}{x}$. Then $(f\circ g\circ h)(x)=\frac{4}{\ln(x^2+e)}$. 2. Since $x^2\ge0$, $x^2+e\ge e$, so $\ln(x^2+e)\ge1$. 3. The denominator takes every value in $[1, \infty)$. Therefore, $k(x)$ is positive, has maximum value $4$, and approaches $0$ without reaching it. The range is $(0, 4]$.

Answer

a) One possible decomposition is $h(x)=x^2+e$, $g(x)=\ln(x)$, and $f(x)=\frac{4}{x}$. b) $(0, 4]$

Function composition and decomposition 34

The functions $a$ and $b$ are given by the table. <table> <tr> <td>$x$</td> <td>$-3$</td> <td>$-2$</td> <td>$-1$</td> <td>$0$</td> <td>$1$</td> </tr> <tr> <td>$a(x)$</td> <td>$0$</td> <td>$-1$</td> <td>$1$</td> <td>$-3$</td> <td>$-2$</td> </tr> <tr> <td>$b(x)$</td> <td>$-1$</td> <td>$1$</td> <td>$0$</td> <td>$-3$</td> <td>$-2$</td> </tr> </table> Create a value table for $f(x)=(a\circ b)(x)$ and $g(x)=(b\circ a)(x)$.

Hints

- In $(a\circ b)(x)$, evaluate $b$ first. - Use the first output as the input for the second table lookup.

Solution

1. For $f(x)=a(b(x))$, read the value of $b(x)$ first and then use it as the input of $a$. This gives $f(-3)=1$, $f(-2)=-2$, $f(-1)=-3$, $f(0)=0$, and $f(1)=-1$. 2. For $g(x)=b(a(x))$, read the value of $a(x)$ first and then use it as the input of $b$. This gives $g(-3)=-3$, $g(-2)=0$, $g(-1)=-2$, $g(0)=-1$, and $g(1)=1$.

Answer

Function composition and decomposition 35

The functions $u$ and $v$ are given by the table. <table> <tr> <td>$x$</td> <td>$1$</td> <td>$2$</td> <td>$3$</td> <td>$4$</td> <td>$5$</td> </tr> <tr> <td>$u(x)$</td> <td>$3$</td> <td>$1$</td> <td>$4$</td> <td>$5$</td> <td>$2$</td> </tr> <tr> <td>$v(x)$</td> <td>$2$</td> <td>$4$</td> <td>$1$</td> <td>$3$</td> <td>$5$</td> </tr> </table> a) Evaluate $u(v(4))$. b) Find all $x\in\{1, 2, 3, 4, 5\}$ for which $(u\circ v)(x)=(v\circ u)(x)$.

Hints

- Evaluate the inner function first. - In part b, compute both compositions for each listed input and compare them.

Solution

1. Since $v(4)=3$ and $u(3)=4$, $u(v(4))=4$. 2. Compare the two compositions at each input. They are equal at $x=1$, where both values are $1$, and at $x=3$, where both values are $3$. 3. At $x=2$, $x=4$, and $x=5$, the two composition values are different. Therefore, the solution set is $\{1, 3\}$.

Answer

a) $4$ b) $x\in\{1, 3\}$

Function composition and decomposition 36

Consider $f(x)=\sqrt{e^{2x}+1}$. a) Find functions $u$ and $v$ such that $f=u\circ v$. b) Decompose $f$ into three functions $p$, $q$, and $r$ such that $f(x)=p(q(r(x)))$. c) Let $g(x)=\frac{1}{e^{2x}+1}$. Using your function $v$ from part a, find a function $w$ such that $g=w\circ v$.

Hints

- Treat the entire radicand as one expression in part a. - Break the inner expression into smaller steps for part b. - In part c, replace the repeated inner expression with a temporary variable.

Solution

1. For part a, choose $v(x)=e^{2x}+1$ and $u(x)=\sqrt{x}$. 2. One three-function decomposition is $r(x)=2x$, $q(x)=e^x+1$, and $p(x)=\sqrt{x}$. 3. Since $v(x)=e^{2x}+1$ and $g(x)=\frac{1}{v(x)}$, choose $w(x)=\frac{1}{x}$.

Answer

a) $v(x)=e^{2x}+1$, $u(x)=\sqrt{x}$ b) One possible answer is $r(x)=2x$, $q(x)=e^x+1$, and $p(x)=\sqrt{x}$. c) $w(x)=\frac{1}{x}$

Function composition and decomposition 37

Let $p$ and $q$ be polynomial functions of degrees $n$ and $m$, respectively, where $n,m\ge1$. 1) Find the degree of $f(x)=(p\circ q)(x)$. Justify your answer using the leading terms. 2) Eleni claims, “Composing two quadratic functions always produces another quadratic function.” Determine whether the claim is true. 3) Let $p(x)=x^2-2$. Find $h(x)=(p\circ p)(x)$ and state its degree.

Hints

- Focus on what substitution does to the highest-degree term of each polynomial. - Use the degree information to test the quadratic claim before expanding a specific example. - For the final composition, substitute the entire expression $p(x)$ into the input of $p$.

Solution

1. Write the leading terms as $p(x)=a_nx^n+\cdots$ and $q(x)=b_mx^m+\cdots$, where $a_n\ne0$ and $b_m\ne0$. The leading term of $p(q(x))$ is $a_n(b_mx^m)^n=a_nb_m^nx^{mn}$. Therefore, the degree is $mn$. 2. Eleni's claim is false. If both functions have degree $2$, their composition has degree $2\cdot2=4$, not degree $2$. 3. $h(x)=(x^2-2)^2-2=x^4-4x^2+2$, so its degree is $4$.

Answer

1) $mn$ 2) The claim is false; the composition has degree $4$. 3) $h(x)=x^4-4x^2+2$, degree $4$

Function composition and decomposition 38

Let $v(x)=2+\frac{1}{x^2+1}$ and $u(t)=\ln(t)$, so $f(x)=(u\circ v)(x)$. a) Find the range of the inner function $v$. b) Use this range and the domain of $u$ to find the domain of $f$. c) Use the fact that $\ln(t)$ is increasing to find the range of $f$.

Hints

- Start by thinking about how small and how large $\frac{1}{x^2+1}$ can be. - Does the fraction ever become exactly zero? - Are all outputs of the inner function allowed inputs for the logarithm? - Which endpoints of the inner range are actually attained?

Solution

1. Since $x^2\ge0$, $0<\frac{1}{x^2+1}\le1$. The value $1$ occurs at $x=0$, but $0$ is approached without being reached. Thus the range of $v$ is $(2,3]$. 2. The logarithm requires a positive input, and all values of $v$ are greater than $2$. Therefore the domain of $f$ is $\mathbb{R}$. 3. The natural logarithm is increasing, so it maps the range $(2,3]$ onto $(\ln(2),\ln(3)]$.

Answer

a) $(2,3]$ b) $\mathbb{R}$ c) $(\ln(2),\ln(3)]$

Function composition and decomposition 39

Let $u(x)=\frac{3}{x-4}$ and $v(x)=x^2+1$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- Substitute the entire inner function into the outer function. - A denominator cannot be zero. - Check the inner function domain before considering the outer function.

Solution

1. $f(x)=u(v(x))=\frac{3}{(x^2+1)-4}=\frac{3}{x^2-3}$. The denominator is zero at $x=\pm\sqrt{3}$, so $D_f=\mathbb{R}\setminus\{-\sqrt{3}, \sqrt{3}\}$. 2. $g(x)=v(u(x))=\left(\frac{3}{x-4}\right)^2+1=\frac{9}{(x-4)^2}+1$. The inner function is undefined at $x=4$, so $D_g=\mathbb{R}\setminus\{4\}$.

Answer

$f(x)=\frac{3}{x^2-3}$, $D_f=\mathbb{R}\setminus\{-\sqrt{3}, \sqrt{3}\}$ $g(x)=\frac{9}{(x-4)^2}+1$, $D_g=\mathbb{R}\setminus\{4\}$

Function composition and decomposition 40

Let $u(x)=\ln(x)$ and $v(x)=4-x^2$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- The input of a logarithm must be positive. - Check the inner function domain first. - A polynomial outer function accepts every real input.

Solution

1. $f(x)=u(v(x))=\ln(4-x^2)$. The logarithm requires $4-x^2>0$, so $-2<x<2$. Therefore, $D_f=(-2, 2)$. 2. $g(x)=v(u(x))=4-(\ln(x))^2$. The logarithm requires $x>0$, while the polynomial outer function adds no restrictions. Therefore, $D_g=(0, \infty)$.

Answer

$f(x)=\ln(4-x^2)$, $D_f=(-2, 2)$ $g(x)=4-(\ln(x))^2$, $D_g=(0, \infty)$

Function composition and decomposition 41

Let $u(x)=\sqrt{x+2}$ and $v(x)=\frac{2}{x-1}$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- Substitute in the correct order. - A square-root radicand must be nonnegative. - A denominator cannot be zero. - Use a sign chart for the rational radicand.

Solution

1. $f(x)=u(v(x))=\sqrt{\frac{2}{x-1}+2}=\sqrt{\frac{2x}{x-1}}$. The radicand must be nonnegative and $x\ne1$. A sign analysis gives $D_f=(-\infty, 0]\cup(1, \infty)$. 2. $g(x)=v(u(x))=\frac{2}{\sqrt{x+2}-1}$. The square root requires $x\ge-2$, and the denominator requires $x\ne-1$. Thus $D_g=[-2, -1)\cup(-1, \infty)$.

Answer

$f(x)=\sqrt{\frac{2x}{x-1}}$, $D_f=(-\infty, 0]\cup(1, \infty)$ $g(x)=\frac{2}{\sqrt{x+2}-1}$, $D_g=[-2, -1)\cup(-1, \infty)$

Function composition and decomposition 42

Let $u(x)=\sqrt{x}$, $v(x)=x^2+2$, and $w(x)=\frac{1}{x}$. a) Find equations and maximal real domains for $g(x)=u(v(w(x)))$ and $h(x)=w(v(u(x)))$. b) Evaluate $g(1)$ and $h(1)$. Explain briefly why the values are different.

Hints

- Work from the inside outward. - Check the reciprocal and square-root restrictions when finding each domain. - Substitute $x=1$ only after simplifying the compositions.

Solution

1. $g(x)=u(v(w(x)))=\sqrt{\frac{1}{x^2}+2}$. The reciprocal requires $x\ne0$, and the radicand is positive for every such $x$. Therefore, $D_g=\mathbb{R}\setminus\{0\}$. 2. $h(x)=w(v(u(x)))=\frac{1}{(\sqrt{x})^2+2}=\frac{1}{x+2}$. The original composition requires $x\ge0$, so $D_h=[0, \infty)$. 3. $g(1)=\sqrt{3}$ and $h(1)=\frac{1}{3}$. The values differ because changing the order of function composition generally changes the result.

Answer

a) $g(x)=\sqrt{\frac{1}{x^2}+2}$, $D_g=\mathbb{R}\setminus\{0\}$; $h(x)=\frac{1}{x+2}$, $D_h=[0, \infty)$ b) $g(1)=\sqrt{3}$ and $h(1)=\frac{1}{3}$

Function composition and decomposition 43

The outer function is $u(x)=\sqrt{x}$, with domain $[0, \infty)$. For each inner function $v$, find the largest subset of its stated domain on which $u(v(x))$ is defined. a) $v(x)=5-2x$, $D_v=\mathbb{R}$ b) $v(x)=x^2-16$, $D_v=\mathbb{R}$ c) $v(x)=\frac{x+4}{x-2}$, $D_v=\mathbb{R}\setminus\{2\}$

Hints

- The output of $v$ must be nonnegative. - Write and solve an inequality for each part. - Use a sign chart for the rational inequality. - Preserve any exclusions from the original domain.

Solution

1. Each composition requires $v(x)\ge0$. 2. For part a, $5-2x\ge0$ gives $x\le\frac{5}{2}$. The domain is $(-\infty, \frac{5}{2}]$. 3. For part b, $x^2-16\ge0$ gives $x\le-4$ or $x\ge4$. The domain is $(-\infty, -4]\cup[4, \infty)$. 4. For part c, solve $\frac{x+4}{x-2}\ge0$ using the critical values $-4$ and $2$. The solution is $x\le-4$ or $x>2$. Thus the domain is $(-\infty, -4]\cup(2, \infty)$.

Answer

a) $(-\infty, \frac{5}{2}]$ b) $(-\infty, -4]\cup[4, \infty)$ c) $(-\infty, -4]\cup(2, \infty)$

Function composition and decomposition 44

Let the outer function be $u(x)=\ln(x)$, whose domain is $(0, \infty)$. For each inner function $v$, find the largest subset of its stated domain on which $u(v(x))$ is defined. a) $v(x)=e^x-5$, $D_v=\mathbb{R}$ b) $v(x)=\frac{1}{x^2-1}$, $D_v=\mathbb{R}\setminus\{-1, 1\}$

Hints

- The input of a logarithm must be positive. - Use logarithms to solve the exponential inequality. - A fraction with positive numerator is positive when its denominator is positive.

Solution

1. Each composition requires $v(x)>0$. 2. For part a, $e^x-5>0$ gives $e^x>5$, so $x>\ln(5)$. The domain is $(\ln(5), \infty)$. 3. For part b, $\frac{1}{x^2-1}>0$. Since the numerator is positive, the denominator must be positive: $x^2-1>0$. Thus $x<-1$ or $x>1$, and the domain is $(-\infty, -1)\cup(1, \infty)$.

Answer

a) $(\ln(5), \infty)$ b) $(-\infty, -1)\cup(1, \infty)$

Function composition and decomposition 45

The figure shows the graphs of a quadratic function $u$ and a linear function $v$. Define $f(x)=u(v(x))$ and $g(x)=v(u(x))$. a) Use the graphs to find $f(0)$ and $g(2)$. Explain the order in which you read each value. b) Use the graphs to find all real zeros of $f$ and $g$. c) Determine equations for $u$ and $v$ from the graphs, and then find equations for $f$ and $g$.

Hints

- Evaluate the inner graph before the outer graph. - For a composition to equal zero, the inner output must be a zero of the outer function. - Use the vertex and intercepts to identify the quadratic equation. - Use two points to identify the linear equation.

Solution

1. Since $v(0)=0$ and $u(0)=-1$, $f(0)=u(v(0))=-1$. Since $u(2)=0$ and $v(0)=0$, $g(2)=v(u(2))=0$. 2. For $f(x)=0$, the value $v(x)$ must equal a zero of $u$. The graph of $u$ has zeros $-2$ and $2$. Since $v(-4)=-2$ and $v(4)=2$, the zeros of $f$ are $-4$ and $4$. 3. For $g(x)=0$, $u(x)$ must equal the zero of $v$, which is $0$. Thus the zeros of $g$ are the zeros of $u$: $-2$ and $2$. 4. The graphs give $u(x)=0.25x^2-1$ and $v(x)=0.5x$. Therefore, $f(x)=0.25(0.5x)^2-1=0.0625x^2-1$, and $g(x)=0.5(0.25x^2-1)=0.125x^2-0.5$.

Answer

a) $f(0)=-1$ and $g(2)=0$ b) Zeros of $f$: $x=-4, 4$; zeros of $g$: $x=-2, 2$ c) $u(x)=0.25x^2-1$, $v(x)=0.5x$, $f(x)=0.0625x^2-1$, and $g(x)=0.125x^2-0.5$

Function composition and decomposition 46

Let $f(x)=x^2$, $g(x)=|x|$, and $h(x)=x-1$. Three of these four compositions are graphed: $u(x)=g(h(x))$, $v(x)=h(g(x))$, $w(x)=f(h(x))$, and $z(x)=h(f(x))$. Match panels p, q, and r to the compositions. Explain how you know.

Hints

- Find the rule for each composition by applying the inner function first. - Where is the corner or vertex of each graph? - Two rules produce a V shape, while the other two produce parabolas.

Solution

1. The compositions are $u(x)=|x-1|$, $v(x)=|x|-1$, $w(x)=(x-1)^2$, and $z(x)=x^2-1$. 2. Graph p is V-shaped with vertex $(1, 0)$, so it represents $u(x)=|x-1|$. 3. Graph q is V-shaped with vertex $(0, -1)$, so it represents $v(x)=|x|-1$. 4. Graph r is a parabola with vertex $(0, -1)$, so it represents $z(x)=x^2-1$. The unused composition is $w(x)=(x-1)^2$.

Answer

p: $u(x)=|x-1|$ q: $v(x)=|x|-1$ r: $z(x)=x^2-1$ Unused: $w(x)=(x-1)^2$

Function composition and decomposition 47

Let $u(x)=e^x$, $v(x)=x^2$, and $w(x)=-x+2$. The figure shows graphs p and q, each produced by composing two of these functions. Identify each composition and justify your choice using symmetry, intercepts, or end behavior.

Hints

- Use symmetry and the vertex to identify the parabola. - Compare y-intercepts and horizontal asymptotes for the exponential graph. - Write equations for likely compositions and compare their features.

Solution

1. Graph p is a downward-opening parabola with vertex $(0, 2)$ and y-axis symmetry. The composition $w(v(x))=-x^2+2$ has exactly these features, so graph p represents $w\circ v$. 2. Graph q approaches the horizontal asymptote $y=2$ as $x\to-\infty$, passes through $(0, 1)$, and decreases without bound as $x\to\infty$. The composition $w(u(x))=-e^x+2$ has these features, so graph q represents $w\circ u$.

Answer

$p: (w\circ v)(x)=-x^2+2$; $q: (w\circ u)(x)=-e^x+2$

Function composition and decomposition 48

The figure shows the graph of $f$. Define $g(x)=e^{f(x)}$. a) Use the graph to find the exact values of $g(0)$ and $g(2)$. b) Determine the end behavior of $g(x)$ as $x\to\infty$ and as $x\to-\infty$. c) Use the graph of $f$ to determine where $g$ reaches its maximum and give the coordinates of the maximum point.

Hints

- Read the inner-function values from the graph first. - Consider what happens to $e^u$ as $u\to-\infty$. - A strictly increasing outer function preserves the location of maxima.

Solution

1. The graph gives $f(0)=1$ and $f(2)=0$. Therefore, $g(0)=e^1=e$ and $g(2)=e^0=1$. 2. The graph of $f$ is a downward-opening parabola, so $f(x)\to-\infty$ as $x\to\pm\infty$. Therefore, $g(x)=e^{f(x)}\to0$ in both directions. 3. Because the exponential function is strictly increasing, $g$ reaches its maximum at the same x-value as $f$. The graph of $f$ has maximum value $1$ at $x=0$, so the maximum point of $g$ is $(0, e)$.

Answer

a) $g(0)=e$ and $g(2)=1$ b) $\lim_{x\to\infty}g(x)=0$ and $\lim_{x\to-\infty}g(x)=0$ c) Maximum point: $(0, e)$

Function composition and decomposition 49

The figure shows the graph of a rational function $f$. For $h(x)=\ln(f(x))$, find the maximal real domain and describe the monotonic behavior of $h$.

Hints

- The input of a logarithm must be positive. - Identify where the graph of $f$ lies above the x-axis. - If two positive outputs of $f$ are in order, what happens to their order after taking logarithms?

Solution

1. The logarithm requires $f(x)>0$. From the graph, $f$ is positive only for $-1<x<3$. The point $x=-1$ is a vertical asymptote and $x=3$ is a zero, so neither endpoint is included. Thus $D_h=(-1, 3)$. 2. The natural logarithm is strictly increasing, so it preserves the monotonic behavior of positive inputs. The graph of $f$ is strictly decreasing on $(-1, 3)$. Therefore, $h$ is strictly decreasing throughout its domain.

Answer

Domain: $(-1, 3)$ Monotonic behavior: strictly decreasing on $(-1, 3)$

Function composition and decomposition 50

The figure shows the graphs of $u(x)=2^x$ and $v(x)=x-2$. Define $f=u\circ v$ and $g=v\circ u$. a) Use the graphs to find $f(3)$ and $g(1)$. b) Use the graphs to solve $f(x)=1$ and $g(x)=0$. c) Describe the transformation that maps the graph of $u$ to the graph of $f$, and the transformation that maps the graph of $u$ to the graph of $g$.

Hints

- Evaluate the inner function first. - Work backward from the output of the outer function when solving an equation. - Changes inside the exponent produce horizontal shifts; changes outside produce vertical shifts.

Solution

1. Since $v(3)=1$ and $u(1)=2$, $f(3)=2$. Since $u(1)=2$ and $v(2)=0$, $g(1)=0$. 2. The equation $f(x)=1$ means $u(v(x))=1$. Since $u(0)=1$, solve $v(x)=0$, which gives $x=2$. The equation $g(x)=0$ means $v(u(x))=0$. Since $v(2)=0$, solve $u(x)=2$, which gives $x=1$. 3. Since $f(x)=2^{x-2}$, its graph is the graph of $u$ shifted $2$ units right. Since $g(x)=2^x-2$, its graph is the graph of $u$ shifted $2$ units down.

Answer

a) $f(3)=2$ and $g(1)=0$ b) $f(x)=1$ at $x=2$; $g(x)=0$ at $x=1$ c) $f$: shift $2$ units right; $g$: shift $2$ units down

Function composition and decomposition 51

The figure shows four graphs created by composing pairs of these functions: $f(x)=2^x$, $g(x)=-x$, and $h(x)=x-3$. Match graphs p–s to the following compositions: $k_a(x)=f(g(x))$, $k_b(x)=g(f(x))$, $k_c(x)=f(h(x))$, and $k_d(x)=h(f(x))$.

Hints

- Write an equation for each composition. - Identify reflections and horizontal or vertical shifts. - Compare y-intercepts and end behavior.

Solution

1. $f(g(x))=2^{-x}$ is a decreasing exponential function with y-intercept $1$, so it is graph p. 2. $g(f(x))=-2^x$ is the graph of $2^x$ reflected across the x-axis, so it is graph q. 3. $f(h(x))=2^{x-3}$ is the graph of $2^x$ shifted $3$ units right, so it is graph r. 4. $h(f(x))=2^x-3$ is the graph of $2^x$ shifted $3$ units down, so it is graph s.

Answer

p: $k_a(x)=f(g(x))$; q: $k_b(x)=g(f(x))$; r: $k_c(x)=f(h(x))$; s: $k_d(x)=h(f(x))$

Function composition and decomposition 52

Let $f(x)=|x|$, $g(x)=x^2-1$, and $h(x)=x-2$. Match graphs p–s to these compositions: $u_1(x)=f(g(x))$, $u_2(x)=g(f(x))$, $u_3(x)=h(f(x))$, and $u_4(x)=h(g(x))$.

Hints

- Evaluate the inner function in each composition before simplifying. - Compare which graphs can take negative values and where they cross the x-axis. - Check whether a graph has a sharp corner or a curved vertex.

Solution

1. $u_1(x)=|x^2-1|$ is never negative and has zeros at $x=-1$ and $x=1$, matching graph p. 2. $u_2(x)=(|x|)^2-1=x^2-1$ is a parabola with vertex $(0, -1)$, matching graph q. 3. $u_3(x)=|x|-2$ is a V-shaped graph with vertex $(0, -2)$, matching graph r. 4. $u_4(x)=(x^2-1)-2=x^2-3$ is a parabola with vertex $(0, -3)$, matching graph s.

Answer

p: $u_1(x)=|x^2-1|$ q: $u_2(x)=x^2-1$ r: $u_3(x)=|x|-2$ s: $u_4(x)=x^2-3$

Function composition and decomposition 53

The coordinate plane shows the graph of a linear function $f$. Define $g(x)=\ln(f(x))$. a) Find the domain of $g$. b) Describe the behavior of $g$ at both ends of its domain. c) Explain whether $g$ is increasing or decreasing on its domain.

Hints

- Identify where the graph of $f$ lies above the x-axis. - Recall the behavior of $\ln(u)$ as $u\to0^+$ and as $u\to\infty$. - Consider the composition of two increasing functions.

Solution

1. From the graph, $f(x)=0.5x+1$. The logarithm requires $f(x)>0$, so $0.5x+1>0$, which gives $x>-2$. Thus, the domain is $(-2, \infty)$. 2. As $x\to-2^+$, $f(x)\to0^+$, so $g(x)\to-\infty$. As $x\to\infty$, $f(x)\to\infty$, so $g(x)\to\infty$. 3. The linear function $f$ is strictly increasing, and $\ln(x)$ is strictly increasing. Therefore, their composition $g$ is strictly increasing on $(-2, \infty)$.

Answer

a) $(-2, \infty)$ b) $\lim_{x\to-2^+}g(x)=-\infty$ and $\lim_{x\to\infty}g(x)=\infty$ c) $g$ is strictly increasing.

Function composition and decomposition 54

The graph shows a rational function $f$. For the composite function $g(x)=\ln(f(x))$, find the domain and describe the monotonicity on each interval of the domain.

Hints

- Find where the graph of $f$ is above the x-axis. - Read the monotonicity of $f$ on those intervals. - An increasing outer function preserves the monotonicity of the inner function.

Solution

1. The logarithm requires $f(x)>0$. From the graph, $f(x)>0$ for $x<-1$ and $x>1$. Thus, the domain is $(-\infty, -1)\cup(1, \infty)$. 2. The graph shows that $f$ is strictly decreasing on both $(-\infty, -1)$ and $(1, \infty)$. 3. Since the natural logarithm is strictly increasing, composing with it preserves the direction of monotonicity. Therefore, $g$ is strictly decreasing on each domain interval.

Answer

Domain: $(-\infty, -1)\cup(1, \infty)$ The function is strictly decreasing on $(-\infty, -1)$ and on $(1, \infty)$.

Function composition and decomposition 55

Let $f(u)=\sqrt{u+2}$. Find a function $g$, defined for $x\ge-4$, such that $f(g(x))=x+4$. Verify your function by composition and explain why the restriction $x\ge-4$ matters in the verification.

Hints

- Replace the input of $f$ by the unknown expression $g(x)$. - Work backward from the required output while respecting the square root's nonnegative output. - In the verification, pay attention to the difference between $\sqrt{z^2}$ and $z$ for arbitrary real $z$.

Solution

1. The equation $f(g(x))=x+4$ means $\sqrt{g(x)+2}=x+4$. 2. For $x\ge-4$, the right side is nonnegative, so squaring gives $g(x)+2=(x+4)^2$. 3. Therefore, $g(x)=(x+4)^2-2$. 4. Verify: $f(g(x))=\sqrt{(x+4)^2}=|x+4|$. Because $x\ge-4$, $|x+4|=x+4$, as required.

Answer

$g(x)=(x+4)^2-2$ for $x\ge-4$. The restriction ensures $\sqrt{(x+4)^2}=|x+4|=x+4$.

Function composition and decomposition 56

Let $f(x)=x^2+1$ and $g(x)=x-k$, where $k>0$. The graph shows part of $h(x)=f(g(x))$. Use its y-intercept and the two function rules to find $k$.

Hints

- What value of $h(0)$ can you read from the graph? - Find $g(0)$ before putting that value into $f$. - How does the condition $k>0$ help choose between possible values?

Solution

1. The graph gives $h(0)=5$. 2. Since $g(0)=-k$, composing the rules gives $h(0)=f(-k)=k^2+1$. 3. Thus $k^2+1=5$, so $k^2=4$. Because $k>0$, we get $k=2$.

Answer

$k=2$, since the graph gives $h(0)=5$ and $f(g(0))=k^2+1$.

Function composition and decomposition 57

Let $u(x)=\ln(x^2-1)$ and $v(x)=e^{x+1}$. Find equations for $f=u\circ v$ and $g=v\circ u$, including the maximal real domain of each composition.

Hints

- Apply the domain restriction of the logarithm. - Use exponential and logarithm properties to simplify. - Simplifying a formula does not restore inputs excluded by the original composition.

Solution

1. $f(x)=u(v(x))=\ln((e^{x+1})^2-1)=\ln(e^{2x+2}-1)$. The logarithm requires $e^{2x+2}-1>0$, which is equivalent to $x>-1$. Thus $D_f=(-1, \infty)$. 2. $g(x)=v(u(x))=e^{\ln(x^2-1)+1}=e(x^2-1)$. The original composition still requires $x^2-1>0$, even though the formula simplifies. Therefore, $D_g=(-\infty, -1)\cup(1, \infty)$.

Answer

$f(x)=\ln(e^{2x+2}-1)$, $D_f=(-1, \infty)$ $g(x)=e(x^2-1)$, $D_g=(-\infty, -1)\cup(1, \infty)$

Function composition and decomposition 58

Construct one function of the form $h(x)=ae^{g(x)}+c$, where $g$ is a polynomial and $a,c\in\mathbb{R}$, whose graph has all three properties: - symmetry about the y-axis, - horizontal asymptote $y=1$, and - a local minimum at $(0,-1)$. State your choices of $a$, $c$, and $g(x)$, and verify the properties.

Hints

- The vertical shift determines the horizontal asymptote. - Choose an even polynomial exponent for y-axis symmetry. - Make the exponent approach $-\infty$ at both ends. - Use the given point to determine the remaining coefficient.

Solution

1. Choose $c=1$ so that $y=1$ can be the horizontal asymptote. 2. Choose the even polynomial $g(x)=-x^2$. Then $g(x)\to-\infty$ as $x\to\pm\infty$, so $e^{g(x)}\to0$, and the resulting function is even. 3. Since $g(0)=0$, the condition $h(0)=-1$ gives $a+1=-1$, so $a=-2$. 4. One valid function is $h(x)=-2e^{-x^2}+1$. It is even, approaches $1$ as $x\to\pm\infty$, and satisfies $h(0)=-1$. 5. Since $e^{-x^2}\le1$ for every real $x$, $-2e^{-x^2}+1\ge-1$, with equality at $x=0$. Therefore, $(0,-1)$ is a local minimum (in fact, a global minimum).

Answer

One valid choice is $a=-2$, $c=1$, and $g(x)=-x^2$, giving $h(x)=-2e^{-x^2}+1$.

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Algebra II math worksheets

Function composition and decomposition

Learning goals

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution

Answer

Hints

Solution