What is the extreme f(x)=x^{1/3}+2x^{4/3} ?

The extreme f(x)=x^{1/3}+2x^{4/3} is Minimum(-1/8 ,-1/2+2(-1/8)^{4/3})

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extreme f(x)=x^{1/3}+2x^{4/3}

extreme points

f (x) = x^{\frac{1}{3}} + 2 x^{\frac{4}{3}}

Minimum (- \frac{1}{8}, - \frac{1}{2} + 2 (- \frac{1}{8})^{\frac{4}{3}})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

: - \frac{1}{8} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{8}

into

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

Minimum (- \frac{1}{8}, - \frac{1}{2} + 2 (- \frac{1}{8})^{\frac{4}{3}})

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

f (x) = \frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4

- 3 (x + 5)^{3}

f (x) = x^{3} - 6 x^{2} + 9 x + 8

f (x) = 2 x + cos (2 x), 0 < x < π

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