Chapter 5.4, Problem 73AYU

In Problems 73-88, use the given function $f$.

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 $f^{-1}$.

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

To determine

To find:

a) The domain of $f$.

Answer to Problem 73AYU

a) The domain is $\left(0,\infty \right)$.

Explanation of Solution

Given:

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

Calculation:

a) The domain of $f$ consists of all $x$ for which $x>0$. Therefore domain is $\left(0, \infty \right)$.

To determine

To find:

b) Graph $f$.

Answer to Problem 73AYU

b)

Explanation of Solution

Given:

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

Calculation:

b) Graph $f$.

$x$	$f\left(x\right)=2+\ln x$
$0.5$	$1.31$
1	2
2	$2.69$

To determine

To find:

c) Range and asymptotes of $f$.

Answer to Problem 73AYU

c) Range of $f$ is $\left(-\infty ,\infty \right)$

Asymptotes of $f$ is $x=\text{ 0}$

Explanation of Solution

Given:

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

Calculation:

c) Range of $f$ is $\left(-\infty , \infty \right)$

Asymptotes of $f$ is $x=0$

To determine

To find:

d) $f^{-1}$

Answer to Problem 73AYU

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

Explanation of Solution

Given:

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

Calculation:

d) To find $f^{-1}$.

Given $y=2+\ln \left(x\right)$

Solve for $y$, $x=2+\ln \left(y\right)$

Taking exponential on both sides $e^x=e^{2+\ln \left(y\right)}$

$e^x=e^2{e}^{ \ln \left(y\right)}$

$e^x=e^{2}y$

$y=\frac{e^x}{e^2}$

$y=e^{x-2}$

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

To determine

To find:

e) Domain and range of $f^{-1}$.

Answer to Problem 73AYU

e) The domain of $f^{-1}$ equal the range of $f=\left(-\infty ,\infty \right)$

The range of $f^{-1}$ equal the domain of $f=\left(0,\infty \right)$

Explanation of Solution

Given:

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

Calculation:

e) The domain of $f^{-1}$ equal the range of $f=\left(-\infty , \infty \right)$

The range of $f^{-1}$ equal the domain of $f=\left(0, \infty \right)$

To determine

To find:

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

Answer to Problem 73AYU

f)

Explanation of Solution

Given:

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

Calculation:

f)

$x$	$f^{-1}\left(x\right)=e^x+3$
$0$	$0.13$
1	$0.36$
$-1$	$0.04$