What is the critical y=x^{4/3}-x^{1/3} ?

The critical y=x^{4/3}-x^{1/3} is x=0,x= 1/4

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critical y=x^{4/3}-x^{1/3}

critical points

y = x^{\frac{4}{3}} - x^{\frac{1}{3}}

x = 0, x = \frac{1}{4}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{1}{4}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{4}{3}} - x^{\frac{1}{3}} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{1}{4}

2 x^{2} + ln (x)

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

f (x, y) = 25 - x^{2} - y^{2}

f (x) = x^{4} - 4 x^{3}, - 1 \leq x \leq 4

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

What is the critical y=x^{4/3}-x^{1/3} ?
The critical y=x^{4/3}-x^{1/3} is x=0,x= 1/4