What is the domain of f(x)=sqrt((x^2-1)/(x^2-x-6)) ?

The domain of f(x)=sqrt((x^2-1)/(x^2-x-6)) is x 3

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domain of f(x)=sqrt((x^2-1)/(x^2-x-6))

domain

f (x) = \frac{x ^{2} - 1}{x ^{2} - x - 6}

x < - 2 or - 1 \leq x \leq 1 or x > 3

Interval Notation

(- \infty, - 2) \cup [- 1, 1] \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x < - 2 or - 1 \leq x \leq 1 or x > 3

Find undefined (singularity) points:

x = 3, x = - 2

Combine real regions and undefined points for final function domain

x < - 2 or - 1 \leq x \leq 1 or x > 3

f (x) = \frac{1}{x ^{2} - 3 x - 10}

x^{3} + 3 x^{2} + 3 x + 1

f (x) = \frac{12 x}{( x ^{2} - 4 ) ^{2}}

f (x) = \frac{x}{x ^{2} + 15 x + 54}

y = x^{4}

What is the domain of f(x)=sqrt((x^2-1)/(x^2-x-6)) ?
The domain of f(x)=sqrt((x^2-1)/(x^2-x-6)) is x<-2\lor-1<= x<= 1\lor x>3