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

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

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

extreme points

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

Minimum (8, - 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 8

into

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

Minimum (8, - 4)

f (x) = x^{2} - 6 x, - 1 \leq x \leq 6

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

\frac{4 x - 12}{( x - 2 ) ^{2}}

f (x) = 3 x^{3} - 3 x^{2} - 3 x + 2, - 1 \leq x \leq 0

f (x, y) = x^{2} + y^{2} - 4 x + 17 x - 10 y + 2021

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