# What are the absolute extrema of #f(x)=x^(1/3)*(20-x)in[0,20]#?

##### 1 Answer

The absolute minimum is

The absolute maximum is

#### Explanation:

The possible points that could be absolute extrema are:

Turning points; i.e. points where

#dy/dx = 0# The endpoints of the interval

We already have our endpoints (

#f'(x) = 0#

#d/dx(x^(1/3)(20-x)) = 0#

#1/3x^(-2/3)(20-x) - x^(1/3) = 0#

#(20-x)/(3x^(2/3)) = x^(1/3)#

#(20-x)/(3x) = 1#

#20-x = 3x#

#20 = 4x#

#5 = x#

So there is a turning point where

#x = 0" "" "x=5" "" "x=20#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's plug these values into

#f(0) = (0)^(1/3) (20 - 0) = 0 * 20 = color(red)0#

#f(5) = (5)^(1/3) (20 - 5) = root(3)(5) * 15 = color(red)(15root(3)5#

#f(20) = (20)^(1/3) (20-20) = root(3)(20) * 0 = color(red)0#

Therefore, on the interval

The absolute minimum is

#color(red)0# , which occurs at#x = 0# and#x=20# .The absolute maximum is

#color(red)(15root(3)5)# , which occurs at#x = 5# .

*Final Answer*