# How do you find the local max and min for  y = 3x^4 + 4x^3 – 12x^2 + 1?

Mar 6, 2016

Overall maximum is 1 at x = 0..
Local minimum $- 4$ at x =1 and $- 31$ at x = $- 2$.

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 \left({x}^{3} + {x}^{2} - 2 x\right) = 12 x \left({x}^{2} + x - 2\right) = 12 x \left(x - 1\right) \left(x + 2\right)$.
= 0, at x 0, x=1 and x=$- 2$.
The second derivative is $12 \left(3 {x}^{2} + 2 x - 2\right)$
This is < 0 at x = 0 and > 0 at x = 1 and x = $- 2$.
Accordingly, y is the maximum at x = 0 and local minimum, at each of x - 1 and x =$- 2$.