# 009A Sample Final 1, Problem 10

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Consider the following continuous function:

$f(x)=x^{1/3}(x-8)$ defined on the closed, bounded interval  $[-8,8]$ .

(a) Find all the critical points for  $f(x)$ .

(b) Determine the absolute maximum and absolute minimum values for  $f(x)$ on the interval  $[-8,8]$ .

Foundations:
1. To find the critical points for  $f(x),$ we set  $f'(x)=0$ and solve for  $x.$ Also, we include the values of  $x$ where  $f'(x)$ is undefined.

2. To find the absolute maximum and minimum of  $f(x)$ on an interval  $[a,b],$ we need to compare the  $y$ values of our critical points with  $f(a)$ and  $f(b).$ Solution:

(a)

Step 1:
To find the critical points, first we need to find  $f'(x).$ Using the Product Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{3}}x^{-{\frac {2}{3}}}(x-8)+x^{\frac {1}{3}}}\\&&\\&=&\displaystyle {{\frac {x-8}{3x^{\frac {2}{3}}}}+x^{\frac {1}{3}}.}\\\end{array}}$ Step 2:
Notice  $f'(x)$ is undefined when  $x=0.$ Now, we need to set  $f'(x)=0.$ So, we get

$-x^{\frac {1}{3}}\,=\,{\frac {x-8}{3x^{\frac {2}{3}}}}.$ We cross multiply to get  $-3x=x-8.$ Solving, we get  $x=2.$ Thus, the critical points for  $f(x)$ are  $(0,0)$ and  $(2,2^{\frac {1}{3}}(-6)).$ (b)

Step 1:
We need to compare the values of  $f(x)$ at the critical points and at the endpoints of the interval.
Using the equation given, we have  $f(-8)=32$ and  $f(8)=0.$ Step 2:
Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for  $f(x)$ is  $32$ and the absolute minimum value for  $f(x)$ is  $2^{\frac {1}{3}}(-6).$ (a)    $(0,0)$ and  $(2,2^{\frac {1}{3}}(-6))$ (b)    The absolute maximum value for  $f(x)$ is  $32$ and the absolute minimum value for  $f(x)$ is  $2^{\frac {1}{3}}(-6).$ 