# Math Help - Critical Value Problem

1. ## Critical Value Problem

Hey all, I'm a little new here and am having trouble toward the end of the semester with my calculus assignments and appreciate all the help I can get.

Find the critical values for the function.
f(x)= 20x^3 – 3x^5

2. To solve this you need to compute the derivative of $f(x)$ and set it equal to 0.

See section 3 for a quick fix.

Derivative - Wikipedia, the free encyclopedia

So for your function we get...

$f(x) = 20x^3 - 3x^5$

$f'(x) = 60x^2 - 15x^4 = 0$

Clearly x=0 is a solution but we also have...

(Divide through by 15)...

$4x^2 - x^4 = 0$

$=> x^2(4 - x^2) = 0$

Divide through by $x^2$...

$4 - x^2 = 0$ which is the same as solving...

$x^2 - 4 = 0$

=> $(x+2)(x-2) = 0$

Hence $x=\pm 2$

So 3 solutions are x=0, x=-2 and x=2.