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

The extreme f(x)=x^3-4x^2+4x is Maximum(2/3 , 32/27),Minimum(2,0)

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

extreme points

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

Maximum (\frac{2}{3}, \frac{32}{27}), Minimum (2, 0)

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 : \frac{32}{27}

Maximum (\frac{2}{3}, \frac{32}{27})

Plug

x = 2

into

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

Minimum (2, 0)

Maximum (\frac{2}{3}, \frac{32}{27}), Minimum (2, 0)

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

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x y + 4

f (x, y) = x^{2} + x y + y^{2} - 16 y + 85

f (t) = (t - 2) u (t + 5)

f (x) = sin^{2} (x), 0 \leq x \leq 2 π

What is the extreme f(x)=x^3-4x^2+4x ?
The extreme f(x)=x^3-4x^2+4x is Maximum(2/3 , 32/27),Minimum(2,0)