What is the extreme (x^3)/3-(x^2)/2-2x+1/3 ?

The extreme (x^3)/3-(x^2)/2-2x+1/3 is Maximum(-1, 3/2),Minimum(2,-3)

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extreme (x^3)/3-(x^2)/2-2x+1/3

extreme

Maximum (- 1, \frac{3}{2}), Minimum (2, - 3)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + \frac{1}{3} : \frac{3}{2}

Maximum (- 1, \frac{3}{2})

Plug

x = 2

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x + \frac{1}{3} : - 3

Minimum (2, - 3)

Maximum (- 1, \frac{3}{2}), Minimum (2, - 3)

What is the extreme (x^3)/3-(x^2)/2-2x+1/3 ?
The extreme (x^3)/3-(x^2)/2-2x+1/3 is Maximum(-1, 3/2),Minimum(2,-3)