Chapter 5, Problem 61RE

(a)

To determine

To Find: The domain of the $f$ .

Answer to Problem 61RE

The domain of the function is $\left(-3,\infty \right)$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$ .

Calculation:

Consider the given function is,

  $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$

The function is defined for the value $\left(x+3\right)>0$ or equivalently $x>-3$ . Then the domain of the function is,

  $\left(-3,\infty \right)$

(b)

To determine

To Find: The graph for the function $f$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$ .

Calculation:

The graph for the given function is shown in Figure 1

  

Figure 1

(c)

To determine

To Find: The range and the asymptote of the graph.

Answer to Problem 61RE

The range of the function is, $\left(-\infty ,\infty \right)$ and the vertical asymptotes of the given function is $x=-3$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$ .

Calculation:

Consider the given function,

  $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$

The range of the above function is the set of all the real numbers that is,

  $\left(-\infty ,\infty \right)$

Thus, the verticalasymptotes of the given function is,

  $x=-3$

(d)

To determine

To Find: The function $f^{-1}$ .

Answer to Problem 61RE

The inverse of the functions is $y=e^{2x}-3$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$ .

Calculation:

Consider the given function,

  $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$

Then,

  $y=\frac{1}{2}\ln \left(x+3\right)$

Then, the inverse of the function is,

  $\begin{array}{l}x=\frac{1}{2}\ln \left(y+3\right)\\ 2x=\ln \left(y+3\right)\\ y=e^{2x}-3\end{array}$

(e)

To determine

To Find: The range of the function.

Answer to Problem 61RE

The range of the function is $\left(-3,\infty \right)$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$ .

Calculation:

Consider the given function,

  $f\left(x\right)=\frac{1}{2}\ln \left(x+3\right)$

The domain of the function $f\left(x^{-1}\right)$ is equal to the range of the $f\left(x\right)$ that is the set of all the real number $\left(-\infty ,\infty \right)$ and the range of the inverse function is the domain of the original function that is $\left(-3,\infty \right)$ in the interval notation.

(f)

To determine

To Find: The graph of the function $f^{-1}$ .

Explanation of Solution

Given:

The inverse of the function is $y=e^{2x}-3$ .

Calculation:

The graph for the inverse of the function is shown in Figure 2

  

Figure 2