Chapter 1.5, Problem 93E

(a)

To determine

To verify: whether $h(x)=\frac{1}{{(x-2)}^2}$ is equal to $f(g(x))$ or not

Explanation of Solution

Given information: $h(x)=\frac{1}{{(x-2)}^2},f(x)=x^2,g(x)=\frac{1}{x-2}$

Proof:

Use $f(x)=x^2,g(x)=\frac{1}{x-2}$ to find $f(g(x))$

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

Also, its given that $h(x)=\frac{1}{{(x-2)}^2}$

So,

  $f(g(x))=h(x)$

(b)

To determine

To verify: whether $h(x)=\frac{1}{{(x-2)}^2}$ is equal to $f(g(x))$ or not

Explanation of Solution

Given information: $h(x)=\frac{1}{{(x-2)}^2},f(x)=\frac{1}{x-2},g(x)=x^2$

Proof:

Use $f(x)=\frac{1}{x-2},g(x)=x^2$ to find $f(g(x))$

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

Also, it’s given that $h(x)=\frac{1}{{(x-2)}^2}$

So,

  $f(g(x))\ne h\left(x\right)$

(c)

To determine

To verify: whether $h(x)=\frac{1}{{(x-2)}^2}$ is equal to $f(g(x))$ or not

Explanation of Solution

Given information: $h(x)=\frac{1}{{(x-2)}^2},f(x)=\frac{1}{x},g(x)={\left(x-2\right)}^2$

Proof:

Use $f(x)=\frac{1}{x},g(x)={\left(x-2\right)}^2$ to find $f(g(x))$

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

Also, it’s given that $h(x)=\frac{1}{{(x-2)}^2}$

So,

  $f(g(x))=h(x)$