What is the extreme (x-8x^2)^{1/3} ?

The extreme (x-8x^2)^{1/3} is Maximum(1/16 , 1/(2*4^{1/3)})

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extreme (x-8x^2)^{1/3}

extreme points

(x - 8 x^{2})^{\frac{1}{3}}

Maximum (\frac{1}{16}, \frac{1}{2 \cdot 4 ^{\frac{1}{3}}})

Solution steps

f^{'} (x) = \frac{1 - 16 x}{3 ( x - 8 x ^{2} ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{16},

Decreasing

: \frac{1}{16} < x < \frac{1}{8},

Decreasing

: \frac{1}{8} < x < \infty

Plug

x = \frac{1}{16}

into

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

Maximum (\frac{1}{16}, \frac{1}{2 \cdot 4 ^{\frac{1}{3}}})

f (x) = x 4 - x^{2}, - 2 \leq x \leq 2

x^{2} + 8 x + 18

f (x) = x e^{- \frac{x}{2}}

1 - x + 1 + x + 2 x

f (x) = - \frac{5 e ^{4 x}}{x - 4}

What is the extreme (x-8x^2)^{1/3} ?
The extreme (x-8x^2)^{1/3} is Maximum(1/16 , 1/(2*4^{1/3)})