Chapter 1.9, Problem 48E

(a)

To determine

To find: inverse of the function of f .

Answer to Problem 48E

The inverse of the given function is $f^{-1}(x)=\sqrt{x+2}$ .

Explanation of Solution

Given information:

The function $f(x)=x^2-2,x\le 0$

Calculation:

Let the function $y=f(x)=x^2-2$

  $y=x^2-2$

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

Hence, the inverse of the given function is $f^{-1}(x)=\sqrt{x+2}$ .

(b)

To determine

To sketch: the graphs both the functions f and $f^{-1}.$

Explanation of Solution

Given information:

The function $f(x)=x^2-2$

Graph:

The graph of the functions $f(x)=x^2-2$ and $f^{-1}(x)=\sqrt{x+2}$ is drawn below.

(c)

To determine

To describe: the relation between the graphs of f and $f^{-1}.$

Explanation of Solution

Given information:

The function $f(x)=x^2-2$

Consider the graph of part $(b)$

The graph of the $f^{-1}$ is the reflection of graph f about the line $y=x.$

(d)

To determine

To state: The domains and ranges of the functions f and $f^{-1}$ .

Explanation of Solution

Given information:

The function $f(x)=x^2-2$

Domain is the set of input values for which the function is defined.

Range is the set of defined output values.

From the graphs of part b,

Consider $f(x)=x^2-2$ is

Domain: $x\ge 0$ , Range: $-1\le y<\infty$

Consider $f^{-1}(x)=\sqrt{x+2}$ is

Domain: $-1\le x<\infty$ , Range: $0\le y<\infty$

Hence, the domains of f and $f^{-1}$ are $[0,\infty ),[-1,\infty )$ and ranges of the functions f and $f^{-1}$

are $[-1,\infty ),[0,\infty )$ respectively.