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

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

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Answer 1 · 2016-03-06T05:00:19

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#.

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

1 Answer

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