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

The extreme f(x)=-x^3-6x^2+6 is Minimum(-4,-26),Maximum(0,6)

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

extreme points

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

Minimum (- 4, - 26), Maximum (0, 6)

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} + 6 : - 26

Minimum (- 4, - 26)

Plug

x = 0

into

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

Maximum (0, 6)

Minimum (- 4, - 26), Maximum (0, 6)

\frac{x ^{2} - 1}{x ^{2} + 4}

f (x, y) = y - x

11 x^{2} + 2 \cdot (- 25) x (1 - x) + 24 (1 - x)^{2}

f (x, y) = x^{2} + y^{2} + \frac{16}{x} + \frac{2}{y}

f (x) = - 5 x^{4} - 5 x^{3} - 2

What is the extreme f(x)=-x^3-6x^2+6 ?
The extreme f(x)=-x^3-6x^2+6 is Minimum(-4,-26),Maximum(0,6)