`f(x) = 1 + 3x^2 - 2x^3` Find the local maximum and minimum values of `f` using both the First and Second Derivative Tests. Which method do you prefer?

Expert Answers
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Given f(x)=1+3x^2-2x^3

f'(x)=6x-6x^2

f''(x)=6-12x

(a) Extrema occur at the critical points; in this case where f'(x)=0.

6x-6x^2=0

(6x)(1-x)=0 ==> x=0 or x=1

For x<0 f'(x)<0
For 0<x<1 f'(x)>0
For x>1 f'(x)<0

Thus there is a minimum at x=0 and a maximum at x=1

(b) f''(0)=6>0 ==> 0 is a local minimum.

f''(1)=-6<0 ==> 1 is a local maximum.

The second derivative test requires less algebra/arithmetic.

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