What is the critical f(x)=ln(x^2-1) ?

The critical f(x)=ln(x^2-1) is None

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critical f(x)=ln(x^2-1)

critical points

f (x) = ln (x^{2} - 1)

None

Solution steps

ln (x^{2} - 1)

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 1, x = 1

Identify critical points not in

f (x)

domain

Domain of

ln (x^{2} - 1) : x < - 1 or x > 1

ln (x^{2} - 1)

is not defined at

x = 0, x = - 1, x = 1

, therefore:

None

\frac{1}{- 10 ( \frac{1}{- 5 x - 6} ) + 3}

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

f (x) = x + 2

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

f (x) = 6 + \frac{1}{7 x}