Question

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

Answer 1

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`.