Chapter 5.1, Problem 17E

To determine

To find: theextreme values of the function on the interval and where they occur by using analytic method and find critical points that are not stationary points.

Answer to Problem 17E

the minimum value is $0$ at $x=0$ and maximum value $3^{\frac{2}{5}}$ at $x=-3$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=x^{\frac{2}{5}},-3\le x\le 1$

Calculation: to find the extreme values of function, first find its derivative,

  $\begin{array}{l}f\left(x\right)=x^{\frac{2}{5}}\\ f\text{'}\left(x\right)=-\frac{5}{3}{x}^{\frac{-3}{5}}\text{ [differentiate and simplify]}\end{array}$

At point $x=0$ where $f\text{'}\left(x\right)=0$ , and the point $x=-3\text{ and }x=1$ , the derivative of function $f\text{'}\left(x\right)>0\text{ and }f\text{'}\left(x\right)<0$ ,

Now substitute in function,

  $\begin{array}{l}f\left(x\right)=x^{\frac{2}{5}}\\ f\left(0\right)=0\text{ [substitute }x=0\text{ and simplify]}\\ f\left(-3\right)=3^{\frac{2}{5}}\text{ [substitute }x=-3\text{ and simplify]}\\ f\left(1\right)=1\text{ [substitute }x=1\text{ and simplify]}\end{array}$

Thus, the minimum value is $0$ at $x=0$ and maximum value $3^{\frac{2}{5}}$ at $x=-3$ .