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

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

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

critical points

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

x = 0, x = \frac{4}{7}, x = 2

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{4}{7}, x = 2, x = 0

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 0, x = \frac{4}{7}, x = 2

x^{3} + 3 x^{2} - 24 x

f (x) = x^{3} + 6 x^{2} + 9 x

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

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

ln (x - y) + x^{2} + y

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