What is the extreme f(x)=(x^3)/3-2x^2-12x ?

The extreme f(x)=(x^3)/3-2x^2-12x is Maximum(-2, 40/3),Minimum(6,-72)

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extreme f(x)=(x^3)/3-2x^2-12x

extreme points

f (x) = \frac{x ^{3}}{3} - 2 x^{2} - 12 x

Maximum (- 2, \frac{40}{3}), Minimum (6, - 72)

Solution steps

f^{'} (x) = x^{2} - 4 x - 12

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - 2 x^{2} - 12 x : \frac{40}{3}

Maximum (- 2, \frac{40}{3})

Plug

x = 6

into

\frac{x ^{3}}{3} - 2 x^{2} - 12 x : - 72

Minimum (6, - 72)

Maximum (- 2, \frac{40}{3}), Minimum (6, - 72)

10 x y

f (x, y) = 64 - x^{2} - y^{2}

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 10

x 4 - x

f (x) = x^{2} + 8 x + 19

What is the extreme f(x)=(x^3)/3-2x^2-12x ?
The extreme f(x)=(x^3)/3-2x^2-12x is Maximum(-2, 40/3),Minimum(6,-72)