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

The extreme f(x)=-x^3-3x^2 is Minimum(-2,-4),Maximum(0,0)

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

extreme points

f (x) = - x^{3} - 3 x^{2}

Minimum (- 2, - 4), Maximum (0, 0)

Solution steps

f^{'} (x) = - 3 x^{2} - 6 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

- x^{3} - 3 x^{2} : - 4

Minimum (- 2, - 4)

Plug

x = 0

into

- x^{3} - 3 x^{2} : 0

Maximum (0, 0)

Minimum (- 2, - 4), Maximum (0, 0)

x^{2} - 6 x - y + 11 = 0

f (x) = \frac{x + 8}{5 + x}

- 5 x + 4 y = 0

f (x) = \frac{10 x + 40}{x - 5}

f (x) = sin (\frac{x}{2})

What is the extreme f(x)=-x^3-3x^2 ?
The extreme f(x)=-x^3-3x^2 is Minimum(-2,-4),Maximum(0,0)