# Difference between revisions of "009A Sample Final 1, Problem 10"

Consider the following continuous function:

${\displaystyle f(x)=x^{1/3}(x-8)}$

defined on the closed, bounded interval  ${\displaystyle [-8,8]}$.

(a) Find all the critical points for  ${\displaystyle f(x)}$.

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

Foundations:
1. To find the critical points for  ${\displaystyle f(x),}$  we set  ${\displaystyle f'(x)=0}$  and solve for  ${\displaystyle x.}$

Also, we include the values of  ${\displaystyle x}$  where  ${\displaystyle f'(x)}$  is undefined.

2. To find the absolute maximum and minimum of  ${\displaystyle f(x)}$  on an interval  ${\displaystyle [a,b],}$

we need to compare the  ${\displaystyle y}$  values of our critical points with  ${\displaystyle f(a)}$  and  ${\displaystyle f(b).}$

Solution:

(a)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{3}}x^{-{\frac {2}{3}}}(x-8)+x^{\frac {1}{3}}}\\&&\\&=&\displaystyle {{\frac {x-8}{3x^{\frac {2}{3}}}}+x^{\frac {1}{3}}.}\\\end{array}}}$

Step 2:
Notice  ${\displaystyle f'(x)}$  is undefined when  ${\displaystyle x=0.}$
Now, we need to set  ${\displaystyle f'(x)=0.}$
So, we get

${\displaystyle -x^{\frac {1}{3}}\,=\,{\frac {x-8}{3x^{\frac {2}{3}}}}.}$

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

(b)

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

(a)    ${\displaystyle (0,0)}$  and  ${\displaystyle (2,2^{\frac {1}{3}}(-6))}$
(b)    The absolute maximum value for  ${\displaystyle f(x)}$  is  ${\displaystyle 32}$  and the absolute minimum value for  ${\displaystyle f(x)}$  is  ${\displaystyle 2^{\frac {1}{3}}(-6).}$