What is the extreme (x^3)/3-(3x^2)/2-4x+8 ?

The extreme (x^3)/3-(3x^2)/2-4x+8 is Maximum(-1, 61/6),Minimum(4,-32/3)

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extreme (x^3)/3-(3x^2)/2-4x+8

extreme

Maximum (- 1, \frac{61}{6}), Minimum (4, - \frac{32}{3})

Solution steps

f^{'} (x) = x^{2} - 3 x - 4

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 4 x + 8 : \frac{61}{6}

Maximum (- 1, \frac{61}{6})

Plug

x = 4

into

\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} - 4 x + 8 : - \frac{32}{3}

Minimum (4, - \frac{32}{3})

Maximum (- 1, \frac{61}{6}), Minimum (4, - \frac{32}{3})

What is the extreme (x^3)/3-(3x^2)/2-4x+8 ?
The extreme (x^3)/3-(3x^2)/2-4x+8 is Maximum(-1, 61/6),Minimum(4,-32/3)