Chapter 5.1, Problem 26E

To determine

To find: theextreme values of the function and where they occur.

Answer to Problem 26E

the minimum value is $1$ at $x=0$ .

Explanation of Solution

Given information:

The function is $y=\frac{1}{\sqrt[3]{1-x^2}}$

Calculation: the function is defined for each value of x,

  $\begin{array}{l}1-x^2>0\\ -1<x<1\end{array}$

The function domain is $\left(-1,1\right)$ , so the domain of function has no end points, so the extreme points occur at critical points,

To find the extreme values of function, first find its derivative,

  $\begin{array}{l}y=\frac{1}{\sqrt[3]{1-x^2}}\\ y\text{'}=-\frac{2}{3}\frac{x}{{\left(1-x^2\right)}^{\frac{2}{3}}}\text{ [differentiate and simplify]}\end{array}$

For critical points equal the derivative of function to zero,

  $\begin{array}{l}-\frac{2}{3}\frac{x}{{\left(1-x^2\right)}^{\frac{2}{3}}}=0\\ x=0\text{ [simplify]}\end{array}$

Now substitute in function,

  $\begin{array}{l}y=\frac{1}{\sqrt[3]{1-x^2}}\\ y=\frac{1}{\sqrt[3]{1-{\left(0\right)}^2}}\text{ [substitute }x=0\text{ ]}\\ y=1\text{ [simplify]}\end{array}$

Thus, the minimum value is $1$ at $x=0$ .