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

The extreme f(x)=(x^3)/3-x^2-3x is Maximum(-1, 5/3),Minimum(3,-9)

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

extreme points

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

Maximum (- 1, \frac{5}{3}), Minimum (3, - 9)

Solution steps

f^{'} (x) = x^{2} - 2 x - 3

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - x^{2} - 3 x : \frac{5}{3}

Maximum (- 1, \frac{5}{3})

Plug

x = 3

into

\frac{x ^{3}}{3} - x^{2} - 3 x : - 9

Minimum (3, - 9)

Maximum (- 1, \frac{5}{3}), Minimum (3, - 9)

What is the extreme f(x)=(x^3)/3-x^2-3x ?
The extreme f(x)=(x^3)/3-x^2-3x is Maximum(-1, 5/3),Minimum(3,-9)