# How do you find the maximum value of 3x^2-x^3?

Aug 11, 2016

$4$

#### Explanation:

$y = 3 {x}^{2} - {x}^{3}$

$y ' = 6 x - 3 {x}^{2} = 3 x \left(2 - x\right)$

$y ' = 0 \implies x = 0 , 2$

$y ' ' = 6 - 6 x = 6 \left(1 - x\right)$

$y ' ' \left(0\right) = 6 > 0$ a min

$y ' ' \left(2\right) = - 6 < 0$ ie a max

$y \left(2\right) = 4$, the max value of y

graph{3x^2 - x^3 [-12.66, 12.66, -6.33, 6.33]}