What is the extreme f(x)=-2x^2(x+6) ?

The extreme f(x)=-2x^2(x+6) is Minimum(-4,-64),Maximum(0,0)

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extreme f(x)=-2x^2(x+6)

extreme points

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

Minimum (- 4, - 64), Maximum (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 4

into

- 2 x^{2} (x + 6) : - 64

Minimum (- 4, - 64)

Plug

x = 0

into

- 2 x^{2} (x + 6) : 0

Maximum (0, 0)

Minimum (- 4, - 64), Maximum (0, 0)

f (x, y) = 3 x^{2} - 6 x + 4 y^{2} + 8

f (x) = 2 x^{3} + 11 x^{2} - 8 x - 12

f (x) = 4 x^{2} + 1 - 3 x

f (x) = 75 - x^{2}

f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2, 0 \leq x \leq 3

What is the extreme f(x)=-2x^2(x+6) ?
The extreme f(x)=-2x^2(x+6) is Minimum(-4,-64),Maximum(0,0)