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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{4}{3}

into

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

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

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

f (x, y) = 5 - (x - 3)^{2} - (y + 2)^{2}

- 3 x^{2} + 5 x y - 2 y^{2} + x + y

f (x) = - x^{2} + 6 x + 1, 5 \leq x \leq 8

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

f (x) = x^{3} - 3 x, 0 \leq x \leq 3

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