Question 285713
 The function f given by f(x)=2x^3-3x^2-12x has a relative minimum at ?
f'(x) = 6x^2-6x-12
Solve f'(x) = 0
x^2-x-2 = 0
(x-2)(x+1) = 0
x = 2 or x = -1
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f''(x) = 2x-1
Find f''(2) = 4-1 = 3 which is >0, so relative minimum at (2,-20)
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{{{graph(400,300,-10,10,-25,25,2x^3-3x^2-12x)}}}
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Cheers,
Stan H.

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(A) -1
(B) 0
(C) 2
(D) 3-√105/4
(E) 3+√105/4