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

The critical f(x)=x^{4/5}(x-4)^2 is x=0,x= 8/7 ,x=4

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

critical points

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

x = 0, x = \frac{8}{7}, x = 4

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{8}{7}, x = 4, x = 0

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 0, x = \frac{8}{7}, x = 4

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

4 x - 7 x^{\frac{4}{7}}

y = \frac{x}{5} - 2

f (x) = 2 - 6 x

x^{4} - 6 x^{2}

What is the critical f(x)=x^{4/5}(x-4)^2 ?
The critical f(x)=x^{4/5}(x-4)^2 is x=0,x= 8/7 ,x=4