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

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

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extreme f(x)=(x^3)/3+x^2-8x

extreme points

f (x) = \frac{x ^{3}}{3} + x^{2} - 8 x

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 4

into

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

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

Plug

x = 2

into

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

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

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

- x^{4} y^{2} - x^{4} + 2 x^{3} y + 2 x^{2} y + x^{2} - 2 x - 2

f (x \cdot y) = x y + 8^{x y}

f (x) = x^{2} - 6 x + 12

f (t) = \frac{x}{x + 25} [0.9]

f (x) = - x^{3} + 3 x^{2} + 429 x + 5242

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