Chapter 4.2, Problem 18E

a.

To determine

To find: the local extrema of the given function.

Answer to Problem 18E

The given function has no local extrema.

Explanation of Solution

Given:

  $k\left(x\right)=\frac{1}{x^2}$ .

Calculation:

Consider the given function,

  $k\left(x\right)=\frac{1}{x^2}$

Differentiating with respect to $x$ , it follows that

  $k^{\prime }\left(x\right)=-\frac{2}{x^3}$

Here, the derivative cannot be zero for any real value of $x$ .

The derivative is not defined at $x=0$ .

So, the critical value is $x=0$ .

But the given function is not defined at $x=0$ .

Hence the given function has no local extrema.

b.

To determine

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

Answer to Problem 18E

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

Explanation of Solution

Given:

  $k\left(x\right)=\frac{1}{x^2}$ .

Concept used:

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

Calculation:

Consider the given function,

  $k\left(x\right)=\frac{1}{x^2}$

Differentiating with respect to $x$ , it follows that

  $k^{\prime }\left(x\right)=-\frac{2}{x^3}$

Now the function increases in the interval for which,

  $\begin{array}{c}{k}^{\prime }\left(x\right)>0\\ -\frac{2}{x^3}>0\\ \frac{2}{x^3}<0\\ x<0\end{array}$

Hence the given function increases in the interval $\left(-\infty ,0\right)$ .

c.

To determine

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

Answer to Problem 18E

The given function decreases in the interval $\left(0,\infty \right)$ .

Explanation of Solution

Given:

  $k\left(x\right)=\frac{1}{x^2}$ .

Concept used:

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

Calculation:

Consider the given function,

  $k\left(x\right)=\frac{1}{x^2}$

Differentiating with respect to $x$ , it follows that

  $h^{\prime }\left(x\right)=-\frac{2}{x^2}$

Now the function decreases in the interval for which,

  $\begin{array}{c}{k}^{\prime }\left(x\right)<0\\ -\frac{2}{x^3}<0\\ \frac{2}{x^3}>0\\ x>0\end{array}$

Hence the given function decreases in the interval $\left(0,\infty \right)$ .