Chapter 1.5, Problem 91E

a.

To determine

To calculate:

Pair of function for $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ equals to $f\left(g\left(x\right)\right)$

Answer to Problem 91E

For Pair of function $g\left(x\right)=\frac{1}{x-2}$ and $f\left(x\right)=x^2$ , $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ is equals to $f\left(g\left(x\right)\right)$ .

Explanation of Solution

Give information:

Given:

  $g\left(x\right)=\frac{1}{x-2}$ and $f\left(x\right)=x^2$

Calculation:

Consider the function, $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$

Given is $g\left(x\right)=\frac{1}{x-2}$ and $f\left(x\right)=x^2$

Now By definition of $g\left(x\right)$ ,

  $f\left(g\left(x\right)\right)=f\left(\frac{1}{x-2}\right)$

Now By definition of $f\left(x\right)$ ,

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

Thus, $f\left(g\left(x\right)\right)=h\left(x\right)$ .

b.

To determine

To calculate:

Pair of function for $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ equals to $f\left(g\left(x\right)\right)$

Answer to Problem 91E

For Pair of function $g\left(x\right)=\frac{1}{x-2}$ and $f\left(x\right)=x^2$ , $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ is not equals to $f\left(g\left(x\right)\right)$ .

Explanation of Solution

Give information:

Given:

  $g\left(x\right)=x^2$ and $f\left(x\right)=\frac{1}{x-2}$

Calculation:

Consider the function, $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$

Given is $g\left(x\right)=x^2$ and $f\left(x\right)=\frac{1}{x-2}$

Now By definition of $g\left(x\right)$ ,

  $f\left(g\left(x\right)\right)=f\left(x^2\right)$

Now By definition of $f\left(x\right)$ ,

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

Thus, $f\left(g\left(x\right)\right)\ne h\left(x\right)$ .

c.

To determine

To calculate:

Pair of function for $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ equals to $f\left(g\left(x\right)\right)$

Answer to Problem 91E

For Pair of function $g\left(x\right)={\left(x-2\right)}^2$ and $f\left(x\right)=\frac{1}{x}$ , $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$ is equals to $f\left(g\left(x\right)\right)$ .

Explanation of Solution

Give information:

Given:

  $g\left(x\right)={\left(x-2\right)}^2$ and $f\left(x\right)=\frac{1}{x}$

Calculation:

Consider the function, $h\left(x\right)=\frac{1}{{\left(x-2\right)}^2}$

Given is $g\left(x\right)={\left(x-2\right)}^2$ and $f\left(x\right)=\frac{1}{x}$

Now By definition of $g\left(x\right)$ ,

  $f\left(g\left(x\right)\right)=f\left({\left(x-2\right)}^2\right)$

Now By definition of $f\left(x\right)$ ,

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

Thus, $f\left(g\left(x\right)\right)=h\left(x\right)$ .