What is the critical f(x)=(-8)/(x^{4/3)(6-x)^{5/3}} ?

The critical f(x)=(-8)/(x^{4/3)(6-x)^{5/3}} is x= 8/3

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critical f(x)=(-8)/(x^{4/3)(6-x)^{5/3}}

critical points

f (x) = \frac{- 8}{x ^{\frac{4}{3}} ( 6 - x ) ^{\frac{5}{3}}}

x = \frac{8}{3}

Solution steps

\frac{- 8}{x ^{\frac{4}{3}} ( 6 - x ) ^{\frac{5}{3}}}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{8}{3}, x = 0, x = 6

Identify critical points not in

f (x)

domain

Domain of

\frac{- 8}{x ^{\frac{4}{3}} ( 6 - x ) ^{\frac{5}{3}}} : x < 0 or 0 < x < 6 or x > 6

\frac{- 8}{x ^{\frac{4}{3}} ( 6 - x ) ^{\frac{5}{3}}}

is not defined at

x = 0, x = 6

, therefore:

x = \frac{8}{3}

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

z = x y - 2 x - 1 - 4 y - 1 + 8

C (x) = 2 x + \frac{300000}{x}, 1 \leq x \leq 300

\frac{x + 2}{x - 1}

f (x) = 1 - 3 x + 5 x^{2} - x^{3}

What is the critical f(x)=(-8)/(x^{4/3)(6-x)^{5/3}} ?
The critical f(x)=(-8)/(x^{4/3)(6-x)^{5/3}} is x= 8/3