What is the domain of y=sqrt(1-ln(x^2-1)) ?

The domain of y=sqrt(1-ln(x^2-1)) is -sqrt(e+1)<= x<-1\lor 1<x<= sqrt(e+1)

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domain of y=sqrt(1-ln(x^2-1))

domain

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

- e + 1 \leq x < - 1 or 1 < x \leq e + 1

Interval Notation

[- e + 1, - 1) \cup (1, e + 1]

Solution steps

Find non-negative values for radicals:

- e + 1 \leq x < - 1 or 1 < x \leq e + 1

The function domain

- e + 1 \leq x < - 1 or 1 < x \leq e + 1

f (x) = \frac{( x + 1 ) ( x + 2 )}{x - 3}

f (x) = 4 sin (x) cos (x)

10 - x^{2}

\frac{7 x}{6 x - 5}

(- \frac{1}{x}) \cdot \frac{1 - x}{1 + x} \cdot \frac{x}{x ^{2} - 1}

What is the domain of y=sqrt(1-ln(x^2-1)) ?
The domain of y=sqrt(1-ln(x^2-1)) is -sqrt(e+1)<= x<-1\lor 1<x<= sqrt(e+1)