What is the critical y=((x-1)^{2/3})/(x^2) ?

The critical y=((x-1)^{2/3})/(x^2) is x=1,x= 3/2

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critical y=((x-1)^{2/3})/(x^2)

critical points

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

x = 1, x = \frac{3}{2}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{3}{2}, x = 0, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{( x - 1 ) ^{\frac{2}{3}}}{x ^{2}} : x < 0 or x > 0

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

is not defined at

x = 0

, therefore:

x = 1, x = \frac{3}{2}

f (x, y) = \frac{1}{3} x^{3} - x y^{2} + y^{2}

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

f (x, y) = 1 + (x - y)^{4} + y^{2}

f (x) = x^{2} - 32 x

f (x, y) = x y + \frac{2}{x} + \frac{4}{y}

What is the critical y=((x-1)^{2/3})/(x^2) ?
The critical y=((x-1)^{2/3})/(x^2) is x=1,x= 3/2