Q 83. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 71– 86, use the given function f  to:

(a) Find the domain of f.

(b) Graph f.

(c) From the graph, determine the range and any asymptotes of f.

(d) Find $f^{-1}$, the inverse of f.

(e) Find the domain and the range of$f^{-1}$.

(f) Graph $f^{-1}$

$f\left(x\right)=e^{x+2}-3$

Step-by-Step Solution

Answer

Part (a) $\left(-\infty ,\infty \right)$

Part (b)  

Part (c) Range $\left(-3,\infty \right)$and vertical asymptote  $y=-3$.

Part (d) $f^{-1}\left(x\right)=\ln (x+3)-2$

Part (e) Domain $\left(-\infty ,\infty \right)$and range $\left(-3,\infty \right)$.

Part (f) 

Part (a) Step 1. Given information.

The given function is: $f\left(x\right)=e^{x+2}-3$

Part (a) Step 2. Find the domain of f.

The domain of f  is the set of all real numbers.

Therefore, the domain of the given function is $\left(-\infty ,\infty \right)$ 

Part (b) Step 1. Graph f.

Sketch the graph of f : 

Part (c) Step 1. Determine the range and any asymptotes of f from the graph.

We can see from the graph of f that the range of the function $f\left(x\right)=e^{x+2}-3$ is $\left(-3,\infty \right)$.

The vertical asymptote of the given function is $y=-3$

Part (d) Step 1. Find f - 1 .

$f\left(x\right)=e^{x+2}-3$

For $f^{-1}$ replace $f\left(x\right)$ with $y$.

$$\begin{gathered} x=e^{y+2}-3 \\ {e}^{y+2}=x+3 \\ y+2=\ln \left(x+3\right) \\ y=\ln \left(x+3\right)-2 \end{gathered}$$

Therefore, the inverse of f  is $f^{-1}\left(x\right)=\ln (x+3)-2$.

Part (e) Find the domain and the range of f - 1 .

We know that the domain of a function f(x) is the range of its inverse.

In the same way, the range of a function f(x) is the domain of its inverse.

Therefore, the domain of $f^{-1}$ is $\left(-\infty ,\infty \right)$and its range is $\left(-3,\infty \right)$.

Part (f) Graph f - 1 .

Sketch the graph of  $f^{-1}$.