What is the critical f(x)=x^{2/5}(x-1) ?

The critical f(x)=x^{2/5}(x-1) is x=0,x= 2/7

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critical f(x)=x^{2/5}(x-1)

critical points

f (x) = x^{\frac{2}{5}} (x - 1)

x = 0, x = \frac{2}{7}

Solution steps

x^{\frac{2}{5}} (x - 1)

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{2}{7}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{2}{5}} (x - 1) : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{2}{7}

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

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

f (x) = (x^{2} - 3) e^{- x}

f (x) = - 2 x^{3} + 33 x^{2} - 108 x + 11

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

What is the critical f(x)=x^{2/5}(x-1) ?
The critical f(x)=x^{2/5}(x-1) is x=0,x= 2/7