What is the critical x^{4/3}+4x^{1/3} ?

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

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critical x^{4/3}+4x^{1/3}

critical points

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

x = - 1, x = 0

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 0

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - 1, x = 0

\frac{4}{2 - x}

f (x) = e^{- x} - 1

y - 3 = - 2 (x + 1)

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

y = - lo g_{3} (x)

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