What is the critical (4x)/(3\sqrt[3]{x^2-4)} ?

The critical (4x)/(3\sqrt[3]{x^2-4)} is x=-2sqrt(3),x=2sqrt(3)

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critical (4x)/(3\sqrt[3]{x^2-4)}

critical points

x = - 23, x = 23

Solution steps

Find where

f^{'} (x)

is equal to zero or undefined

x = - 23, x = 23, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

is not defined at

x = - 2, x = 2

, therefore:

x = - 23, x = 23

\frac{5 x - 15}{3 ( x - 1 ) ^{\frac{1}{3}}}

y = \frac{4 x ^{5}}{5} - \frac{5 x ^{4}}{4} + 4 x - 1

f (x, y) = x^{2} + y^{2} + x y - 3 x

y = 1 - \frac{1}{3 t ^{\frac{2}{3}}}

2 x^{3} - 6 x^{2} - 18 x + 54

What is the critical (4x)/(3\sqrt[3]{x^2-4)} ?
The critical (4x)/(3\sqrt[3]{x^2-4)} is x=-2sqrt(3),x=2sqrt(3)