What is the extreme f(x)=8x^{1/3}-x^{4/3} ?

The extreme f(x)=8x^{1/3}-x^{4/3} is Maximum(2,6*2^{1/3})

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extreme f(x)=8x^{1/3}-x^{4/3}

extreme points

f (x) = 8 x^{\frac{1}{3}} - x^{\frac{4}{3}}

Maximum (2, 6 \cdot 2^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{8 - 4 x}{3 x ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 2

into

8 x^{\frac{1}{3}} - x^{\frac{4}{3}} : 6 \cdot 2^{\frac{1}{3}}

Maximum (2, 6 \cdot 2^{\frac{1}{3}})

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

f (x) = x^{2} + 6 x + 15

f (x) = ln (x)

f (x) = - x^{3} - 9 x^{2}

f (x) = - 3

What is the extreme f(x)=8x^{1/3}-x^{4/3} ?
The extreme f(x)=8x^{1/3}-x^{4/3} is Maximum(2,6*2^{1/3})