What is the extreme f(x)=(x^3)/3-(x^2)/2-2x ?

The extreme f(x)=(x^3)/3-(x^2)/2-2x is Maximum(-1, 7/6),Minimum(2,-10/3)

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extreme f(x)=(x^3)/3-(x^2)/2-2x

extreme points

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x

Maximum (- 1, \frac{7}{6}), Minimum (2, - \frac{10}{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{7}{6}

Maximum (- 1, \frac{7}{6})

Plug

x = 2

into

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x : - \frac{10}{3}

Minimum (2, - \frac{10}{3})

Maximum (- 1, \frac{7}{6}), Minimum (2, - \frac{10}{3})

f (x, y) = - x^{2} + x - y^{2} - 2 y

f (x) = 8 x ln (x)

f (x) = x^{3} - 3 x^{2} - 9 x + 11

f (x) = x^{3} - 3 x^{2} - 9 x + 20

f (x) = \frac{2}{3} x^{3} + 3 x^{2} - 8 x + 1

What is the extreme f(x)=(x^3)/3-(x^2)/2-2x ?
The extreme f(x)=(x^3)/3-(x^2)/2-2x is Maximum(-1, 7/6),Minimum(2,-10/3)