# 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:
Recall:
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 minimum value for ${\displaystyle f(x)}$ is ${\displaystyle 2^{\frac {1}{3}}(-6).}$