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(-2, 10/3),Minimum(1,-7/6)

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

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

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

Plug

x = 1

into

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x : - \frac{7}{6}

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

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

14 x^{2} - 2 x^{3} + 2 y^{2} + 4 x y

f (x) = - x^{3} + 12 x - 8

f (x) = x^{2} - 11, - 2 \leq x \leq 3

f (x y) = x^{3} + y^{2} - 6 x y + 9 x + 5 y + 2

f (x, y) = e^{- x - y^{2}} \cdot (3 x - y^{2})

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(-2, 10/3),Minimum(1,-7/6)