What is the extreme f(x)=-x^3-3x^2-1 ?

The extreme f(x)=-x^3-3x^2-1 is Minimum(-2,-5),Maximum(0,-1)

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extreme f(x)=-x^3-3x^2-1

extreme points

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

Minimum (- 2, - 5), Maximum (0, - 1)

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} - 1 : - 5

Minimum (- 2, - 5)

Plug

x = 0

into

- x^{3} - 3 x^{2} - 1 : - 1

Maximum (0, - 1)

Minimum (- 2, - 5), Maximum (0, - 1)

y = - \frac{5}{2} x + 4

8 x + 6 y = - 4

13 = 0 \cdot 16 + b

y = x^{2} + 4

f (x) = \frac{x ^{3} + 6 x ^{2} + 9 x}{x + 3}

What is the extreme f(x)=-x^3-3x^2-1 ?
The extreme f(x)=-x^3-3x^2-1 is Minimum(-2,-5),Maximum(0,-1)