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

1 Answer
Mar 6, 2016

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

Explanation:

dy/dx = 12(x^3+x^2-2 x) = 12 x (x^2+x-2)=12x(x-1)(x+2).
= 0, at x 0, x=1 and x=-2.
The second derivative is 12 (3x^2+2 x-2)
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.