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# Whats the local maximum and minimum points for x^3-2x^2-8x?

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- DaMLv 61 decade agoFavorite Answer
If y=x^3-2x^2-8x

then dy/dx=3x^2-4x-8

equate to 0 and solve for x

3x^2-4x-8=0

using quadratic formula, with a=3, b=-4, and c=-8

we get x=-1.0972 and x=2.4305

plug in those values in the origional formula to get the corresponding values of y

first pair, x=-1.097,y=5.049

2nd pair x=2.431, y=-16.901

EDIT to add,

sahsjing, you factored wrong in

3x^2-4x-8 = (3x+4)(x-2) = 0

(3x+4)(x-2)=3x^2-2x-8

Source(s): some mental, confirmed by both graph and numeric solution on calculator, (and also a computer program) - sahsjingLv 71 decade ago
y = f(x) = x^3-2x^2-8x

f'(x) = 3x^2-4x-8 = (3x+4)(x-2) = 0

Critical points: -4/3, 2

local maximum: f(-4/3) = 128/27

local minimum: f(2) = -16

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