What is the extreme f(x)=-x^3-6x^2+1 ?

The extreme f(x)=-x^3-6x^2+1 is Minimum(-4,-31),Maximum(0,1)

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extreme f(x)=-x^3-6x^2+1

extreme points

f (x) = - x^{3} - 6 x^{2} + 1

Minimum (- 4, - 31), Maximum (0, 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 4

into

- x^{3} - 6 x^{2} + 1 : - 31

Minimum (- 4, - 31)

Plug

x = 0

into

- x^{3} - 6 x^{2} + 1 : 1

Maximum (0, 1)

Minimum (- 4, - 31), Maximum (0, 1)

- 5 x - 12 y = 11

- \frac{2}{3} x - 5

\frac{5}{12 + 3 x}

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

f (x) = 3 x^{6} + x^{10} + 4

What is the extreme f(x)=-x^3-6x^2+1 ?
The extreme f(x)=-x^3-6x^2+1 is Minimum(-4,-31),Maximum(0,1)