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

Question

Christopher Agresta · Accepted Answer

The absolute minimum is #0#, which occurs at #x = 0# and #x=20#.

The absolute maximum is #15root(3)5#, which occurs at #x = 5#.

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 (#0# and #20#), so let's find our turning points: #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 = 5#. This means that the 3 possible points that could be extrema are: #x = 0" "" "x=5" "" "x=20# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Let's plug these values into #f(x)#: #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 #x in [0, 20]#: 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

Isabella Ackerman · Answer

undefined To find the absolute extrema of $ f(x) = x^{1/3} (20 - x) $ on the interval [0, 20], we first evaluate the function at the critical points and endpoints, then compare the values to determine the absolute extrema.

1. Critical points:
To find critical points, we take the derivative of $ f(x) $ and set it equal to zero:
$ f'(x) = \frac{d}{dx} [x^{1/3} (20 - x)] $
$ = \frac{1}{3} x^{-2/3} (20 - x) - x^{1/3} $
Setting $ f'(x) = 0 $ and solving for $ x $:
$ \frac{1}{3} x^{-2/3} (20 - x) - x^{1/3} = 0 $
$ \frac{20 - x}{3x^{2/3}} - x^{1/3} = 0 $
$ \frac{20 - x}{3x^{2/3}} = x^{1/3} $
$ 20 - x = 3x $
$ 20 = 4x $
$ x = 5 $

2. Endpoints:
Evaluate $ f(x) $ at the endpoints of the interval [0, 20]:
$ f(0) = 0 $
$ f(20) = 20^{1/3} (20 - 20) = 0 $

Now, compare the values:
$ f(0) = 0 $
$ f(5) = 5^{1/3} (20 - 5) = 5^{1/3} \times 15 $
$ f(20) = 0 $

The absolute minimum is at $ x = 5 $, and the absolute maximum is at the endpoints $ x = 0 $ and $ x = 20 $.