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
To solve this you need to compute the derivative of $\displaystyle f(x)$ and set it equal to 0.
See section 3 for a quick fix.
Derivative - Wikipedia, the free encyclopedia
So for your function we get...
$\displaystyle f(x) = 20x^3 - 3x^5$
$\displaystyle f'(x) = 60x^2 - 15x^4 = 0$
Clearly x=0 is a solution but we also have...
(Divide through by 15)...
$\displaystyle 4x^2 - x^4 = 0$
$\displaystyle => x^2(4 - x^2) = 0$
Divide through by $\displaystyle x^2$...
$\displaystyle 4 - x^2 = 0$ which is the same as solving...
$\displaystyle x^2 - 4 = 0$
=> $\displaystyle (x+2)(x-2) = 0$
Hence $\displaystyle x=\pm 2$
So 3 solutions are x=0, x=-2 and x=2.