SOLUTION: please help me to answer this question, the directions are:
determine the intervals of increasing and decreasing.
f(x)=-2x^3-3x^2+12
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Quadratic Equations and Parabolas
-> SOLUTION: please help me to answer this question, the directions are:
determine the intervals of increasing and decreasing.
f(x)=-2x^3-3x^2+12
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Question 143597: please help me to answer this question, the directions are:
determine the intervals of increasing and decreasing.
f(x)=-2x^3-3x^2+12 Answer by nabla(475) (Show Source):
You can put this solution on YOUR website! f'(x)=-6x^2-6x
Find where it is=0
0=-6x^2-6x
0=6x(x+1)
implies x=0 or x=-1. These are our two critical values. Find out what happens outside and between these two values for the derivative f'.
f'(-2)=-6(-2)^2-6(-2)=-24+12=-12. It is decreasing from (,-1).
f'(-1/2)=-6(-1/2)^2-6(-1/2)=-3/2+3=3/2. It is increasing from (-1,0).
f(1)=-6-6=-12. It is decreasing from (0,).
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