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

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- Apr 13th 2010, 06:24 AMakuczma86Critical 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 - Apr 13th 2010, 06:46 AMDeadstar
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.