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  $0.$

Final Answer:
(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  $0.$

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