What is the extreme f(x)=(\sqrt[3]{6x^2-x^3}) ?

The extreme f(x)=(\sqrt[3]{6x^2-x^3}) is Minimum(0,0),Maximum(4,2\sqrt[3]{4})

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extreme f(x)=(\sqrt[3]{6x^2-x^3})

extreme points

Solution steps

f^{'} (x) = \frac{x ( - x + 4 )}{( 6 x ^{2} - x ^{3} ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = 0

into

Minimum (0, 0)

Plug

x = 4

into

f (x) = 7 x^{3} - 7 x^{2} + 7

y = - \frac{4 x}{x ^{2} + 1}

f (x) = \frac{- 2}{3} x^{3} - 6 x^{2} - 10 x + 80

\frac{4 x + 5}{x - 7}

g (θ) = 4 θ - 7 sin (θ), 0 \leq θ \leq π

What is the extreme f(x)=(\sqrt[3]{6x^2-x^3}) ?
The extreme f(x)=(\sqrt[3]{6x^2-x^3}) is Minimum(0,0),Maximum(4,2\sqrt[3]{4})