What is the domain of (sqrt(x^2-9))/(x^2+2x-8) ?

The domain of (sqrt(x^2-9))/(x^2+2x-8) is x = 3

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domain of (sqrt(x^2-9))/(x^2+2x-8)

domain

\frac{x ^{2} - 9}{x ^{2} + 2 x - 8}

x < - 4 or - 4 < x \leq - 3 or x \geq 3

Interval Notation

(- \infty, - 4) \cup (- 4, - 3] \cup [3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = 2, x = - 4

Combine real regions and undefined points for final function domain

x < - 4 or - 4 < x \leq - 3 or x \geq 3

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

\frac{5}{x - 4}

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

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

y = \frac{4}{\frac{2}{x - 1} + 3}

What is the domain of (sqrt(x^2-9))/(x^2+2x-8) ?
The domain of (sqrt(x^2-9))/(x^2+2x-8) is x<-4\lor-4<x<=-3\lor x>= 3