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

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

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

extreme points

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

Minimum (- 1, - 3)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

x^{\frac{4}{3}} + 4 x^{\frac{1}{3}} : - 3

Minimum (- 1, - 3)

h (x) = - 3 x^{5} + 5 x^{3}

f (x) = ln (5 - 2 x^{2})

f (x) = 2 x^{3} - \frac{11}{2} x^{2} - 10 x + 2

f (x, y) = 8 y^{2} + 5 x^{2} - 10 y + 6 x - 10

f (x) = 8 x^{5} - 120 x^{3} + 43

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