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

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

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

extreme points

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

Maximum (0, 0), Minimum (\frac{4}{3}, - \frac{32}{27})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{4}{3}

into

x^{3} - 2 x^{2} : - \frac{32}{27}

Minimum (\frac{4}{3}, - \frac{32}{27})

Maximum (0, 0), Minimum (\frac{4}{3}, - \frac{32}{27})

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

f (x) = 5 x + 6

2 x^{2} - 1

\frac{1}{- x + 5}

f (x) = \frac{4}{x + 2} + 1

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