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

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

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

extreme points

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

Minimum (0, - 6), Maximum (2, - 2)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 6)

Plug

x = 2

into

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

Maximum (2, - 2)

Minimum (0, - 6), Maximum (2, - 2)

f (x) = \frac{1}{x - 4} + \frac{1}{6 - x}

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

2 x + y = 12

f (x) = \frac{( x + 1 )}{x ^{2} - 9}

f (x) = 4 + 4 + x

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