Locate all relative extrema using second derivative test: f(x)=4x^3 -27x^2 -30x -4

Can you please show me the steps.  Thanks!
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f'(x)=12x^2-54x-30. Extrema at f'(x)=0, so 6x^2-27x-15=0=(3x-15)(2x+1)=0, and x=5 and -1/2.

f(5)=500-675-150-4=-329 and f(-1/2)=-1/2-27/4+15-4=15/4.

f''(x)=24x-54=6(4x-9), so f''(5)>0 (minimum) and f''(-1/2)<0 (maximum).

The extrema are (5,-329) minimum and (-1/2,15/4) maximum.
by Top Rated User (1.2m points)

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