What is the extreme f(x)=-x^3+3x^2-4 ?

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

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extreme f(x)=-x^3+3x^2-4

extreme points

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

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

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} - 4 : - 4

Minimum (0, - 4)

Plug

x = 2

into

- x^{3} + 3 x^{2} - 4 : 0

Maximum (2, 0)

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

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

f (x) = \frac{( 1 - 2 x )}{5 + x}

y = 3^{x}

f (x) = \frac{4}{x - 3}

f (x) = x (5 - x) (2 x - 3)

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