What is the extreme f(x)=x^3-4x^2+4x-1 ?

The extreme f(x)=x^3-4x^2+4x-1 is Maximum(2/3 , 5/27),Minimum(2,-1)

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extreme f(x)=x^3-4x^2+4x-1

extreme points

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

Maximum (\frac{2}{3}, \frac{5}{27}), Minimum (2, - 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < \frac{2}{3},

Decreasing

: \frac{2}{3} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = \frac{2}{3}

into

x^{3} - 4 x^{2} + 4 x - 1 : \frac{5}{27}

Maximum (\frac{2}{3}, \frac{5}{27})

Plug

x = 2

into

x^{3} - 4 x^{2} + 4 x - 1 : - 1

Minimum (2, - 1)

Maximum (\frac{2}{3}, \frac{5}{27}), Minimum (2, - 1)

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x - y + 10

f (x) = \frac{x ^{4}}{4} - 8 x

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

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

f (x) = 4 x^{3} - 16 x

What is the extreme f(x)=x^3-4x^2+4x-1 ?
The extreme f(x)=x^3-4x^2+4x-1 is Maximum(2/3 , 5/27),Minimum(2,-1)