What is the critical F(x)=x^{4/5}(x-6)^2 ?

The critical F(x)=x^{4/5}(x-6)^2 is x=0,x= 12/7 ,x=6

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critical F(x)=x^{4/5}(x-6)^2

critical points

F (x) = x^{\frac{4}{5}} (x - 6)^{2}

x = 0, x = \frac{12}{7}, x = 6

Solution steps

x^{\frac{4}{5}} (x - 6)^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{12}{7}, x = 6, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{4}{5}} (x - 6)^{2} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{12}{7}, x = 6

cos (6 x), 0 \leq x \leq \frac{π}{2}

\frac{2 x ^{2} + 5}{x ^{2} - 25}

f (x) = - 12 x - 12

\frac{ω ^{2} + 2 ω + 1}{3 ω - 5}

f (x) = x^{2} + 2 x y + 2 y^{2} - 6 x + 10 y + 4

What is the critical F(x)=x^{4/5}(x-6)^2 ?
The critical F(x)=x^{4/5}(x-6)^2 is x=0,x= 12/7 ,x=6