Chapter 1, Problem 9PS

a.

To determine

To find: $(f\circ g)(x)$.

Answer to Problem 9PS

  $(f\circ g)(x)=4x+24$ .

Explanation of Solution

Given Information: Given functions are $f(x)=4xandg(x)=x+6$ .

Calculation:

  $\begin{array}{l}(f\circ g)(x)=f(g(x))\\ =f(x+6)\\ =4(x+6)\\ =4x+24\end{array}$

Thus, $(f\circ g)(x)=4x+24$ .

b.

To determine

To find: ${(f\circ g)}^{-1}(x)$.

Answer to Problem 9PS

  ${(f\circ g)}^{-1}(x)=\frac{x}{6}-4$ .

Explanation of Solution

Given Information: Given functions are $f(x)=4xandg(x)=x+6$ .

Calculation:

From part $(a)$ , $(f\circ g)(x)=4x+24$ .

Let $y=4x+24$ .

Replacing $x$ with $y$ ,

  $x=4y+24$

  $\begin{array}{l}\Rightarrow 4y=x-24\\ \Rightarrow y=\frac{x}{6}-4\end{array}$

Replacing $y$ with ${(f\circ g)}^{-1}(x)$ ,

  ${(f\circ g)}^{-1}(x)=\frac{x}{6}-4$ .

c.

To determine

To find: $f^{-1}(x)and{g}^{-1}(x)$.

Answer to Problem 9PS

  $f^{-1}(x)=\frac{x}{4}and{g}^{-1}(x)=x-6$ .

Explanation of Solution

Given Information: Given functions are $f(x)=4xandg(x)=x+6$ .

Calculation:

  $f(x)=4x$

Let $y=4x$ .

Replacing $x$ with $y$ ,

  $x=4y$

  $\Rightarrow y=\frac{x}{4}$

Thus, $f^{-1}(x)=\frac{x}{4}$ .

  $g(x)=x+6$

Let $y=x+6$ .

  $\Rightarrow x=y-6$

Replacing $x$ with $y$ ,

  $y=x-6$ .

Thus, $g^{-1}(x)=x-6$ .

d.

To determine

To find: $(g^{-1}\circ f^{-1})(x)$ and to compare the result with that of part $(b)$.

Answer to Problem 9PS

  $(g^{-1}\circ f^{-1})(x)=\frac{x}{4}-6={(f\circ g)}^{-1}(x)$ .

Explanation of Solution

Given Information: Given functions are $f(x)=4xandg(x)=x+6$ .

Calculation:

  $\begin{array}{l}(g^{-1}\circ f^{-1})(x)=g^{-1}(f^{-1}(x))\\ =g^{-1}\left(\frac{x}{4}\right)\\ =\frac{x}{4}-6\\ ={(}^f(x)\end{array}$

Thus, $(g^{-1}\circ f^{-1})(x)=\frac{x}{4}-6={(f\circ g)}^{-1}(x)$ .

e.

To determine

To find: $(f\circ g)(x),{(f\circ g)}^{-1}(x),f^{-1}(x),g^{-1}(x),(g^{-1}\circ f^{-1})(x)$

Answer to Problem 9PS

  $(f\circ g)(x)=8x^3+1,{(f\circ g)}^{-1}(x)=\frac{1}{2}\sqrt[3]{x-1},f^{-1}(x)=\sqrt[3]{x-1},g^{-1}(x)=\frac{x}{2},(g^{-1}\circ f^{-1})(x)=\frac{\sqrt[3]{x-1}}{2}={(f\circ g)}^{-1}(x)$

Explanation of Solution

Given Information: Given functions are $f(x)=x^3+1andg(x)=2x$ .

Calculation:

f.

To determine

To find: $(f\circ g)(x),{(f\circ g)}^{-1}(x),f^{-1}(x),g^{-1}(x),(g^{-1}\circ f^{-1})(x)$ for two one-one functions $fandg$ .

Answer to Problem 9PS

For $f(x)=2x+1andg(x)=5x$ ,

  $(f\circ g)(x)=10x+1,{(f\circ g)}^{-1}(x)=\frac{x-1}{10},f^{-1}(x)=\frac{x-1}{2},g^{-1}(x)=\frac{x}{5},(g^{-1}\circ f^{-1})(x)=\frac{x-1}{10}={(f\circ g)}^{-1}(x).$

Explanation of Solution

Given Information: Functions $fandg$ should be one-one.

Calculation:

Let $f(x)=2x+1andg(x)=5x$ .

Then $fandg$ are one-one functions.

1. $(f\circ g)(x)=f(g(x))=f(5x)=2(5x)+1=10x+1$
2. $f(x)=2x+1$

Let $y=10x+1$ .

Replacing $x$ with $y$ ,

  $\begin{array}{l}x=10y+1\\ \Rightarrow y=\frac{x-1}{10}\\ \Rightarrow {(}^f(x)=\frac{x-1}{10}\end{array}$

3. $f(x)=2x+1$

Let $y=2x+1$ .

Replacing $x$ with $y$ ,

  $\begin{array}{l}x=2y+1\\ \Rightarrow y=\frac{x-1}{2}\\ \Rightarrow f^{-1}(x)=\frac{x-1}{2}\end{array}$

  $g(x)=5x$

Let $y=5x$ .

Replacing $x$ with $y$ ,

  $\begin{array}{l}x=5y\\ \Rightarrow y=\frac{x}{5}\\ \Rightarrow g^{-1}(x)=\frac{x}{5}\end{array}$

4. $(g^{-1}\circ f^{-1})(x)=g^{-1}(f^{-1}(x))$

  $\begin{array}{l}=g^{-1}\left(\frac{x-1}{2}\right)\\ =\frac{x-1}{10}\\ ={(}^f(x)\end{array}$

So, $(g^{-1}\circ f^{-1})(x)=\frac{x-1}{10}={(f\circ g)}^{-1}(x)$ .

g.

To determine

To make: A conjecture about ${(f\circ g)}^{-1}(x)and(g^{-1}\circ f^{-1})(x).$

Answer to Problem 9PS

  ${(f\circ g)}^{-1}(x)=(g^{-1}\circ f^{-1})(x).$

Explanation of Solution

Given Information: Given functions are $f(x)=4xandg(x)=x+6$ .

From above parts, it can be concluded that ${(f\circ g)}^{-1}(x)=(g^{-1}\circ f^{-1})(x).$