What is the critical 4(x-5)^{2/3} ?

The critical 4(x-5)^{2/3} is x=5

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critical 4(x-5)^{2/3}

critical points

4 (x - 5)^{\frac{2}{3}}

x = 5

Solution steps

4 (x - 5)^{\frac{2}{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 5

Identify critical points not in

f (x)

domain

Domain of

4 (x - 5)^{\frac{2}{3}} : - \infty < x < \infty

All critical points are in domain

x = 5

f (x) = x + 4 - \frac{1 - x}{x}

f (x) = a x^{2}

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

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