What is the domain of f(x)=((x^2+16))/(x^2-9) ?

The domain of f(x)=((x^2+16))/(x^2-9) is x 3

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domain of f(x)=((x^2+16))/(x^2-9)

domain

f (x) = \frac{( x ^{2} + 16 )}{x ^{2} - 9}

x < - 3 or - 3 < x < 3 or x > 3

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 3, x = - 3

The function domain

x < - 3 or - 3 < x < 3 or x > 3

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

f (x) = 2 3 x + 3 + 5

- 2 x + 3 x + 4

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

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

What is the domain of f(x)=((x^2+16))/(x^2-9) ?
The domain of f(x)=((x^2+16))/(x^2-9) is x<-3\lor-3<x<3\lor x>3