How do you find the local max and min for #x^3-2x^2-8x#?

1 Answer
Topscooter
Jan 17, 2016

Take a look at the derivative.

Explanation:

Here, #f(x) = x^3 - 2x^2 - 8x# so #f'(x) = 3x^2 - 4x - 8#.

The local max and min of #f# will be given by the roots of #f'#.

We first calculate #Delta = 16 - 4*3*(-8) = 112# so #f'# has 2 real roots.

By the quadratic formula, the zeros of #f'# are given by #(-b +- sqrtDelta)/2a#, which are here #(4 +-sqrt112)/6#.

So the critical points are #f((4 + sqrt112)/6)# and #f((4 -sqrt112)/6)#. By evaluating them, you will see which one is a local max and which one is the local min.

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