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

The extreme f(x)=(x^3)/3-(x^2)/2-2x+2 is Maximum(-1, 19/6),Minimum(2,-4/3)

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

extreme points

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

Maximum (- 1, \frac{19}{6}), Minimum (2, - \frac{4}{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 + 2 : \frac{19}{6}

Maximum (- 1, \frac{19}{6})

Plug

x = 2

into

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

Minimum (2, - \frac{4}{3})

Maximum (- 1, \frac{19}{6}), Minimum (2, - \frac{4}{3})

s (t) = - t^{2} + 5 t + 1

f (x) = x^{3} + 3 x^{4}

f (x) = - x + 49

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 2 x - 9

f (x) = x^{2} (2 x - 1)^{\frac{1}{3}} [0.1]

What is the extreme f(x)=(x^3)/3-(x^2)/2-2x+2 ?
The extreme f(x)=(x^3)/3-(x^2)/2-2x+2 is Maximum(-1, 19/6),Minimum(2,-4/3)