# 009A Sample Final 3, Problem 9

Let

$g(x)=(2x^{2}-8x)^{\frac {2}{3}}$ (a) Find all critical points of  $g$ over the  $x$ -interval  $[0,8].$ (b) Find absolute maximum and absolute minimum of  $g$ over  $[0,8].$ 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  $g'(x).$ Using the Chain Rule, we have

${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {{\frac {2}{3}}(2x^{2}-8x)^{-{\frac {1}{3}}}(2x^{2}-8x)'}\\&&\\&=&\displaystyle {{\frac {2}{3}}(2x^{2}-8x)^{-{\frac {1}{3}}}(4x-8)}\\&&\\&=&\displaystyle {{\frac {8x-16}{3{\sqrt[{3}]{2x^{2}-8x}}}}.}\end{array}}$ Step 2:
First, we note that  $g'(x)$ is undefined when
$3{\sqrt[{3}]{2x^{2}-8x}}=0.$ Solving for  $x,$ we get
${\begin{array}{rcl}\displaystyle {0}&=&\displaystyle {2x^{2}-8x}\\&&\\&=&\displaystyle {x(2x-8).}\end{array}}$ Therefore,  $g'(x)$ is undefined when  $x=0,4.$ Now, we need to set  $g'(x)=0.$ So, we get

$8x-16=0.$ Solving, we get  $x=2.$ Thus, the critical points for  $f(x)$ are  $(0,0),(2,4),(4,0).$ (b)

Step 1:
We need to compare the values of  $g(x)$ at the critical points and at the endpoints of the interval.
Using the equation given, we have  $g(0)=0$ and  $g(8)=16.$ Step 2:
Comparing the values in Step 1 with the critical points in (a), the absolute maximum value for  $g(x)$ is  $16$ and the absolute minimum value for  $g(x)$ is  $0.$ (a)   $(0,0),(2,4),(4,0).$ (b)   The absolute maximum value for  $g(x)$ is  $16$ and the absolute minimum value for  $g(x)$ is  $0.$ 