Chapter 4.2, Problem 26E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 26E

The local minimum point is $\left(2,-\frac{1}{4}\right)$ and the local maximum point is $\left(-2,\frac{1}{4}\right)$ .

Explanation of Solution

Given:

  $h\left(x\right)=\frac{-x}{x^2+4}$ .

Calculation:

Consider the given function,

  $h\left(x\right)=\frac{-x}{x^2+4}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{h}^{\prime }\left(x\right)=-\left[\frac{1\cdot \left(x^2+4\right)-\left(2x+0\right)\cdot x}{{\left(x^2+4\right)}^2}\right]\\ =-\left[\frac{x^2+4-2x^2}{{\left(x^2+4\right)}^2}\right]\\ =\frac{x^2-4}{{\left(x^2+4\right)}^2}\end{array}$

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

  $\begin{array}{c}{h}^{\prime }\left(x\right)=0\\ \frac{x^2-4}{{\left(x^2+4\right)}^2}=0\\ x^2-4=0\\ x^2=4\\ x=\pm 2\end{array}$

Therefore $x=-2,2$ are the critical values.

The graph of $h\left(x\right)=\frac{-x}{x^2+4}$ rises to the left of $x=-2$ and falls to the right of $x=-2$ .

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

The graph of $h\left(x\right)=\frac{-x}{x^2+4}$ 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,

  $\begin{array}{c}h\left(2\right)=\frac{-2}{2^2+4}\\ =-\frac{1}{4}\end{array}$

And the local maximum value is,

  $\begin{array}{c}h\left(-2\right)=\frac{-\left(-2\right)}{{\left(-2\right)}^2+4}\\ =\frac{1}{4}\end{array}$

b.

To determine

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

Answer to Problem 26E

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

Explanation of Solution

Given:

  $h\left(x\right)=\frac{-x}{x^2+4}$ .

Concept used:

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

Calculation:

Consider the given function,

  $h\left(x\right)=\frac{-x}{x^2+4}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{h}^{\prime }\left(x\right)=-\left[\frac{1\cdot \left(x^2+4\right)-\left(2x+0\right)\cdot x}{{\left(x^2+4\right)}^2}\right]\\ =-\left[\frac{x^2+4-2x^2}{{\left(x^2+4\right)}^2}\right]\\ =\frac{x^2-4}{{\left(x^2+4\right)}^2}\end{array}$

Now the function increases in the interval for which,

  $\begin{array}{c}{h}^{\prime }\left(x\right)>0\\ \frac{x^2-4}{{\left(x^2+4\right)}^2}>0\\ x^2-4>0\\ x^2>4\\ x<-2\text{or}x>2\end{array}$

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

c.

To determine

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

Answer to Problem 26E

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

Explanation of Solution

Given:

  $h\left(x\right)=\frac{-x}{x^2+4}$ .

Concept used:

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

Calculation:

Consider the given function,

  $h\left(x\right)=\frac{-x}{x^2+4}$

Differentiating with respect to $x$ , it follows that

  $\begin{array}{c}{h}^{\prime }\left(x\right)=-\left[\frac{1\cdot \left(x^2+4\right)-\left(2x+0\right)\cdot x}{{\left(x^2+4\right)}^2}\right]\\ =-\left[\frac{x^2+4-2x^2}{{\left(x^2+4\right)}^2}\right]\\ =\frac{x^2-4}{{\left(x^2+4\right)}^2}\end{array}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{h}^{\prime }\left(x\right)<0\\ \frac{x^2-4}{{\left(x^2+4\right)}^2}<0\\ x^2-4<0\\ x^2<4\\ -2<x<-2\end{array}$

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