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 , 59/27),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{59}{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{59}{27}

Maximum (\frac{2}{3}, \frac{59}{27})

Plug

x = 2

into

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

Minimum (2, 1)

Maximum (\frac{2}{3}, \frac{59}{27}), Minimum (2, 1)

f (x) = - 2

f (x) = \frac{x - 6}{x ( x - 7 )}

f (x) = \frac{5}{( x - 3 )}

f (x) = 2 x + 1 - 3

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

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 , 59/27),Minimum(2,1)