What is the extreme f(x)=4x^2-5x^3 ?

The extreme f(x)=4x^2-5x^3 is Minimum(0,0),Maximum(8/15 , 256/675)

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extreme f(x)=4x^2-5x^3

extreme points

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

Minimum (0, 0), Maximum (\frac{8}{15}, \frac{256}{675})

Solution steps

f^{'} (x) = 8 x - 15 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{8}{15},

Decreasing

: \frac{8}{15} < x < \infty

Plug

x = 0

into

4 x^{2} - 5 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{8}{15}

into

4 x^{2} - 5 x^{3} : \frac{256}{675}

Maximum (\frac{8}{15}, \frac{256}{675})

Minimum (0, 0), Maximum (\frac{8}{15}, \frac{256}{675})

x^{3} - 6 x^{2} + 15

\frac{4 x}{25 - x ^{2}}

f (x) = x^{\frac{1}{5}} (x + 6)

f (x) = x^{2} + 0.3 x - 0.22 x

y = x (3 + 5 z)

What is the extreme f(x)=4x^2-5x^3 ?
The extreme f(x)=4x^2-5x^3 is Minimum(0,0),Maximum(8/15 , 256/675)