Chapter 5.2, Problem 18E

a.

To determine

To find the local extrema.

Answer to Problem 18E

The local extrema at none.

Explanation of Solution

Given information:

The given function is $k\left(x\right)=\frac{1}{x^2}$

Concept used:

Extreme values occur only at critical points and end points.

A point in the interior of the domain of a function f at which $f^{\prime }=0$ or $f^{\prime }$ does not exist is called a critical point of f .

Calculation :

Take derivative of the function

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

Equate the derivative to 0.

  $-\frac{2}{x^3}=0$

The value of x is not defined

Therefore,

The local extrema at none.

b.

To determine

To find the intervals on which the function is increasing.

Answer to Problem 18E

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

Explanation of Solution

Given information:

The given function is $k\left(x\right)=\frac{1}{x^2}$

Concept used:

If $f^{\prime }>0$ at each point of $\left(a,b\right)$ , then f increases on $\left[a,b\right]$

It is assumed that f be continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ .

Calculation :

For increasing function $k^{\prime }\left(x\right)$ must be greater than 0.

  $\begin{array}{l}-\frac{2}{x^3}>0\\ \frac{-2}{x}\cdot {\left(\frac{1}{x}\right)}^2>0\end{array}$

  $k^{\prime }\left(x\right)$ Has two factors one is a perfect square which is always non-negative and other is multiplied by a negative sign that means when x is positive $k^{\prime }\left(x\right)$ is negative and when x is negative $k^{\prime }\left(x\right)$ is positive. At $x=0$

  $k^{\prime }\left(x\right)$ is undefined.

Therefore,

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

c.

To determine

To find the intervals on which the function is decreasing.

Answer to Problem 18E

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

Explanation of Solution

Given information:

The given function is $k\left(x\right)=\frac{1}{x^2}$

Concept used:

If $f^{\prime }<0$ at each point of $\left(a,b\right)$ , then f decreases on $\left[a,b\right]$

It is assumed that f be continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ .

Calculation :

For decreasing function $k^{\prime }\left(x\right)$ must be less than 0.

  $\begin{array}{l}-\frac{2}{x^3}<0\\ \frac{-2}{x}\cdot {\left(\frac{1}{x}\right)}^2<0\end{array}$

  $k^{\prime }\left(x\right)$ Has two factors one is a perfect square which is always non-negative and other is multiplied by a negative sign that means when x is positive $k^{\prime }\left(x\right)$ is negative and when x is negative $k^{\prime }\left(x\right)$ is positive. At $x=0$

  $k^{\prime }\left(x\right)$ is undefined.

Therefore,

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