Chapter 4.3, Problem 30E

(a)

To determine

To find:The intervals of increase or decrease.

Explanation of Solution

Given: $f\left(x\right)=\ln \left(x^4+27\right)$

Calculate the first derivative of the given function.

  $f^{\prime }\left(x\right)=\frac{4x^3}{x^4+27}$

Equate to zero.

  $\begin{array}{c}0=\frac{4x^3}{x^4+27}\\ x=0\end{array}$

The function is decreasing in the interval $\left(-\infty ,0\right)$ .

The function is increasing in the interval $\left(0,\infty \right)$

(b)

To determine

To find :The local maxima and local minima.

Explanation of Solution

The function is decreasing in the interval $\left(-\infty ,0\right)$ .

The function is increasing in the interval $\left(0,\infty \right)$

The value of $f^{\prime }$ is changing from negative to positive at $x=0$

So, the local minima is $\left(0,\ln 27\right)$

There is no local maxima.

(c)

To determine

To find :The concavity of the function and the point of inflection.

Explanation of Solution

Calculate the second derivative of the given function.

  $\begin{array}{c}{f}^{″}\left(x\right)=\frac{324x^2-4x^6}{{\left(x^4+27\right)}^2}\\ =\frac{4x^2\left(x^4-81\right)}{{\left(x^4+27\right)}^2}\\ 4x^2=0\\ x=0\end{array}$

  $f^{″}\left(x\right)<0$ for the interval $-\infty <x<-3$ . So, the graph is concave down for $\left(-\infty ,-3\right)$

  $f^{″}\left(x\right)>0$ for the interval $-3<x<0$ . So, the graph is concave up for $\left(-3,0\right)$

  $f^{″}\left(x\right)>0$ for the interval $0<x<3$ . So, the graph is concave up for $\left(0,3\right)$

  $f^{″}\left(x\right)<0$ for the interval $3<x<\infty$ . So, the graph is concave down for $\left(3,\infty \right)$

The inflection points are $\left(-3.4.68\right),\left(0,3.30\right)\text{and}\left(3,4.68\right)$

(d)

To determine

To sketch :The graph of result obtained from part (a) to (c).

Explanation of Solution

Given: $f\left(x\right)=\ln \left(x^4+27\right)$

The graph of the result obtained is shown in figure below.

  

Figure (1)

Therefore, the required graph is shown in figure (1).