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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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})

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)