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

The extreme f(x)= 2/3 x^3-2x^2 is Maximum(0,0),Minimum(2,-8/3)

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extreme f(x)= 2/3 x^3-2x^2

extreme points

f (x) = \frac{2}{3} x^{3} - 2 x^{2}

Maximum (0, 0), Minimum (2, - \frac{8}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

\frac{2}{3} x^{3} - 2 x^{2} : 0

Maximum (0, 0)

Plug

x = 2

into

\frac{2}{3} x^{3} - 2 x^{2} : - \frac{8}{3}

Minimum (2, - \frac{8}{3})

Maximum (0, 0), Minimum (2, - \frac{8}{3})

f (x) = x^{4} - 5 x^{3} + 9 x^{2}

f (x, y) = x y - 2 x - y + 6

f (x) = - (x - 1)^{2}

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

f (x, y) = \frac{1}{3} x^{2} + 2 x y + 3 y^{2} + 4 x - 1

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