What is the critical f(x)=ln(4-x^2) ?

The critical f(x)=ln(4-x^2) is x=0

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critical f(x)=ln(4-x^2)

critical points

f (x) = ln (4 - x^{2})

x = 0

Solution steps

ln (4 - x^{2})

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

ln (4 - x^{2}) : - 2 < x < 2

ln (4 - x^{2})

is not defined at

x = - 2, x = 2

, therefore:

x = 0

f (x, y) = y x - y^{2} - 3 x + 11 y

y = 3 x^{2} - 4 x + 7

f (x, y) = x^{2} y - x^{2} - \frac{1}{3} y^{3}

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

f (x) = 2 + x + \frac{7}{x}