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

The extreme f(x)=-x^3+4x^2 is Minimum(0,0),Maximum(8/3 , 256/27)

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

extreme points

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

Minimum (0, 0), Maximum (\frac{8}{3}, \frac{256}{27})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{8}{3},

Decreasing

: \frac{8}{3} < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{8}{3}

into

- x^{3} + 4 x^{2} : \frac{256}{27}

Maximum (\frac{8}{3}, \frac{256}{27})

Minimum (0, 0), Maximum (\frac{8}{3}, \frac{256}{27})

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

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

f (x, y) = 3 x y - 5 x - 2 y + 1

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

f (x) = - 2 x^{2} - 2 y^{2} + 20 x + 16 y + 4

What is the extreme f(x)=-x^3+4x^2 ?
The extreme f(x)=-x^3+4x^2 is Minimum(0,0),Maximum(8/3 , 256/27)