What is the extreme f(x)=((x^3))/3-2x^2-5x ?

The extreme f(x)=((x^3))/3-2x^2-5x is Maximum(-1, 8/3),Minimum(5,-100/3)

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extreme f(x)=((x^3))/3-2x^2-5x

extreme points

f (x) = \frac{( x ^{3} )}{3} - 2 x^{2} - 5 x

Maximum (- 1, \frac{8}{3}), Minimum (5, - \frac{100}{3})

Solution steps

f^{'} (x) = x^{2} - 4 x - 5

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - 2 x^{2} - 5 x : \frac{8}{3}

Maximum (- 1, \frac{8}{3})

Plug

x = 5

into

\frac{x ^{3}}{3} - 2 x^{2} - 5 x : - \frac{100}{3}

Minimum (5, - \frac{100}{3})

Maximum (- 1, \frac{8}{3}), Minimum (5, - \frac{100}{3})

What is the extreme f(x)=((x^3))/3-2x^2-5x ?
The extreme f(x)=((x^3))/3-2x^2-5x is Maximum(-1, 8/3),Minimum(5,-100/3)