Chapter 4.2, Problem 24E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 24E

The local minimum point is $\left(-2,6{\left(-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)$ and there is no local maximum.

Explanation of Solution

Given:

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$ .

Calculation:

Consider the given function,

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{1}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(x+8\right)+\left(1+0\right)\cdot x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8+3x}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ =\frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\end{array}$

Now to find the critical values, solve the first derivative by equating it to zero. That is,

  $\begin{array}{c}{g}^{\prime }\left(x\right)=0\\ \frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=0\\ 4x+8=0\\ 4x=-8\\ x=-2\end{array}$

Therefore $x=-2$ is the critical value.

The graph of $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$ falls to the left of $x=-2$ and rises to the right of $x=-2$ .

Therefore $x=-2$ is the point of local minima.

Now the local minimum value is,

  $\begin{array}{c}g\left(-2\right)={\left(-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(-2+8\right)\\ =6{\left(-2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\end{array}$

b.

To determine

To find: the intervals in which the given function is increasing.

Answer to Problem 24E

The given function increases in the interval $\left(-2,\infty \right)$ .

Explanation of Solution

Given:

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$ .

Concept used:

The function increases in the interval at which the first derivative is positive.

Calculation:

Consider the given function,

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{1}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(x+8\right)+\left(1+0\right)\cdot x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8+3x}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ =\frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\end{array}$

Now the function increases in the interval for which,

  $\begin{array}{c}{g}^{\prime }\left(x\right)>0\\ \frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}>0\\ 4x+8>0\\ 4x>-8\\ x>-2\end{array}$

Hence the given function increases in the interval $\left(-2,\infty \right)$ .

c.

To determine

To find: the intervals in which the given function is decreasing.

Answer to Problem 24E

The given function decreases in the interval $\left(-\infty ,-2\right)$ .

Explanation of Solution

Given:

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$ .

Concept used:

The function decreases in the interval at which the first derivative is negative.

Calculation:

Consider the given function,

  $g\left(x\right)=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\left(x+8\right)$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{g}^{\prime }\left(x\right)=\frac{1}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\left(x+8\right)+\left(1+0\right)\cdot x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}+x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ =\frac{x+8+3x}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\\ =\frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\end{array}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{g}^{\prime }\left(x\right)<0\\ \frac{4x+8}{3x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}<0\\ 4x+8<0\\ 4x<-8\\ x<-2\end{array}$

Hence the given function decreases in the interval $\left(-\infty ,-2\right)$ .