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

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