What is the extreme f(x)=-x^4+4x^3-4x^2-8 ?

The extreme f(x)=-x^4+4x^3-4x^2-8 is Maximum(0,-8),Minimum(1,-9),Maximum(2,-8)

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extreme f(x)=-x^4+4x^3-4x^2-8

extreme points

f (x) = - x^{4} + 4 x^{3} - 4 x^{2} - 8

Maximum (0, - 8), Minimum (1, - 9), Maximum (2, - 8)

Solution steps

f^{'} (x) = - 4 x^{3} + 12 x^{2} - 8 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- x^{4} + 4 x^{3} - 4 x^{2} - 8 : - 8

Maximum (0, - 8)

Plug

x = 1

into

- x^{4} + 4 x^{3} - 4 x^{2} - 8 : - 9

Minimum (1, - 9)

Plug

x = 2

into

- x^{4} + 4 x^{3} - 4 x^{2} - 8 : - 8

Maximum (2, - 8)

Maximum (0, - 8), Minimum (1, - 9), Maximum (2, - 8)

\frac{x - 3}{x ^{2} - 4 x + 4}

y = 6 x - 1

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

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

f (x) = 3 x^{2} + 4 x + 1

What is the extreme f(x)=-x^4+4x^3-4x^2-8 ?
The extreme f(x)=-x^4+4x^3-4x^2-8 is Maximum(0,-8),Minimum(1,-9),Maximum(2,-8)