Find Absolute max and min

[mshireling]

Find the absolute maximum and minimum values of the following functions on the given intervals. Calculators are allowed for arithmetic only. Show all work.


f(x)=x[sup:1ohusltw]1/3[/sup:1ohusltw](8-x) Domain[-1,8]


I need help taking the derivative of this function please!

 

[wjm11]

Find the absolute maximum and minimum values of the following functions on the given intervals. Calculators are allowed for arithmetic only. Show all work.


f(x)=x^(1/3)(8-x) Domain[-1,8]


I need help taking the derivative of this function please!

In its current form, you would apply the Product Rule. If you want to use only the Power Rule, expand the equation first:

f(x)=x^(1/3)(8-x)
f(x) = 8x^(1/3) - x^(4/3)

Can you take it from here?

 

[mshireling]

so then the derivative would be (8/3)x[sup:hzmjjxvk](-2/3)[/sup:hzmjjxvk]-(4/3)x[sup:hzmjjxvk](1/3)[/sup:hzmjjxvk]

 

[wjm11]

the derivative would be (8/3)x^(-2/3)-(4/3)x^(1/3)

Correct. Good job.

Next, remember to check the end point values on your closed interval when searching for extrema as well as the places where f' = 0.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ x^{1/3}(8-x) \ = \ 8x^{1/3}-x^{4/3} \ over\ interval \ [-1,8]\)

\(\displaystyle f \ ' \ (x) \ = \ 8(1/3)x^{-2/3}-(4/3)x^{1/3} \ = \ \frac{8}{3x^{2/3}}-\frac{4x^{1/3}}{3} \ = \ \frac{8-4x}{3x^{2/3}}\)

\(\displaystyle Two \ critical \ points, \ x \ = \ 2 \ and \ x \ = \ 0.\)

\(\displaystyle f(-1) \ = \ -9\)

\(\displaystyle f(0) \ = \ 0\)

\(\displaystyle f(2) \ = \ 2^{1/3}*6 \ = \ 7.55.....\)

\(\displaystyle f(8) \ = \ 0\)

\(\displaystyle Hence, \ absolute \ min. \ equals \ [-1,-9] \ and \ absolute \ max \ equals \ [2,7.55...], \ see \ graph \ below.\)

[attachment=0:30h4uwjh]neither.jpg[/attachment:30h4uwjh]