What is the extreme f(x)=(x^3)/3-3x^2-7x ?

The extreme f(x)=(x^3)/3-3x^2-7x is Maximum(-1, 11/3),Minimum(7,-245/3)

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extreme f(x)=(x^3)/3-3x^2-7x

extreme points

f (x) = \frac{x ^{3}}{3} - 3 x^{2} - 7 x

Maximum (- 1, \frac{11}{3}), Minimum (7, - \frac{245}{3})

Solution steps

f^{'} (x) = x^{2} - 6 x - 7

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - 3 x^{2} - 7 x : \frac{11}{3}

Maximum (- 1, \frac{11}{3})

Plug

x = 7

into

\frac{x ^{3}}{3} - 3 x^{2} - 7 x : - \frac{245}{3}

Minimum (7, - \frac{245}{3})

Maximum (- 1, \frac{11}{3}), Minimum (7, - \frac{245}{3})

What is the extreme f(x)=(x^3)/3-3x^2-7x ?
The extreme f(x)=(x^3)/3-3x^2-7x is Maximum(-1, 11/3),Minimum(7,-245/3)