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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq - 4

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

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

f (x) = 3 x^{5} - 5 x^{3} + 3

y = 4 x^{2}

f (x) = 2 cos (6 x)

f (x) = \frac{3}{x ^{2} - 5}

3 x + 3 y = 21

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