What is the domain of f(x)=((sqrt(x^2-9)))/(x+4) ?

The domain of f(x)=((sqrt(x^2-9)))/(x+4) is x = 3

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domain of f(x)=((sqrt(x^2-9)))/(x+4)

domain

f (x) = \frac{( x ^{2} - 9 )}{x + 4}

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 = - 4

Combine real regions and undefined points for final function domain

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

y = - 5 - 3 x - 15 - 6

f (x) = x^{2} + 10 x + 2

f (x) = x^{2} + 10 x + 6

f (x) = - 2 (4) cos (6^{2} x + \frac{π}{4}) - 2

\frac{x}{2 x - 8}

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