What is the extreme y=(x^3)/3-x^2-8x ?

The extreme y=(x^3)/3-x^2-8x is Maximum(-2, 28/3),Minimum(4,-80/3)

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extreme y=(x^3)/3-x^2-8x

extreme points

y = \frac{x ^{3}}{3} - x^{2} - 8 x

Maximum (- 2, \frac{28}{3}), Minimum (4, - \frac{80}{3})

Solution steps

f^{'} (x) = x^{2} - 2 x - 8

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - x^{2} - 8 x : \frac{28}{3}

Maximum (- 2, \frac{28}{3})

Plug

x = 4

into

\frac{x ^{3}}{3} - x^{2} - 8 x : - \frac{80}{3}

Minimum (4, - \frac{80}{3})

Maximum (- 2, \frac{28}{3}), Minimum (4, - \frac{80}{3})

3 x^{2} - 18 x + 24

f (x) = x^{4} - 4 x^{3} + 9, - 5 \leq x \leq 2

f (x, y) = x^{2} y^{2} - 5 x^{2} - 8 x y - 5 y^{2}

\frac{1}{5} x + 1

f (x) = \frac{2 x ^{2} - 2}{x ^{2} - 4}

What is the extreme y=(x^3)/3-x^2-8x ?
The extreme y=(x^3)/3-x^2-8x is Maximum(-2, 28/3),Minimum(4,-80/3)