Q 73. - 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 $f^{-1}$.

$f\left(x\right)=2+\ln x$

Step-by-Step Solution

Answer

Part (a) $\left(0,\infty \right)$.

Part (b) 

Part (c) Range $\left(-\infty ,\infty \right)$ and vertical asymptote $x=0$.

Part (d) $f^{-1}\left(x\right)=e^{x-2}$.

Part (e) Domain $\left(-\infty ,\infty \right)$ and range $\left(0,\infty \right)$.

Part (f) 

Part (a) Step 1. Given information.

The given function is:

$f\left(x\right)=2+\ln x$

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

$f\left(x\right)=2+\ln x$

The domain of f consists of all x for which $x>0$.

Therefore, the domain of the given function f  is $\left(0,\infty \right)$.

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 function $f\left(x\right)=2+\ln x$ is the set of all real numbers.

Therefore, the range of the function is $\left(-\infty ,\infty \right)$.

The vertical asymptote of the given function is $x=0$.

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

$f\left(x\right)=2+\ln x$

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

$$\begin{gathered} y=2+\ln x \\ x=2+\ln y \\ \ln y=x-2 \\ y=e^{x-2} \\ {f}^{-1}\left(x\right)=e^{x-2} \end{gathered}$$

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

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 $f^{-1} \mathrm{is} \left(-\infty ,\infty \right)$ and its range is $\left(0,\infty \right)$.

Part (f) Graph f - 1 .

Sketch the graph of $f^{-1}$: