What is the critical f(x)=x^{4/5}(x-1)^2 ?

The critical f(x)=x^{4/5}(x-1)^2 is x=0,x= 2/7 ,x=1

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critical f(x)=x^{4/5}(x-1)^2

critical points

f (x) = x^{\frac{4}{5}} (x - 1)^{2}

x = 0, x = \frac{2}{7}, x = 1

Solution steps

x^{\frac{4}{5}} (x - 1)^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{2}{7}, x = 1, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{4}{5}} (x - 1)^{2} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{2}{7}, x = 1

f (x) = 80 x^{3} - 3 x^{5}

y = x (1 - x^{2})^{2}

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

f (x) = - 2 x^{2} - 12 x - 13

f (x, y) = y x - y^{2} - x + 6 y

What is the critical f(x)=x^{4/5}(x-1)^2 ?
The critical f(x)=x^{4/5}(x-1)^2 is x=0,x= 2/7 ,x=1