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:

Solution:

(a)

Step 1:
To find the critical point, 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 $2^{\frac {1}{3}}(-6)$.

Final Answer:
(a) $(0,0)$ and $(2,2^{\frac {1}{3}}(-6))$
(b) The absolute minimum value for $f(x)$ is $2^{\frac {1}{3}}(-6)$

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