What is the domain of f(x)=((x^2-3x))/(x^2-9) ?

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

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domain of f(x)=((x^2-3x))/(x^2-9)

domain

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

f (x) = ln (\frac{( 9 - x ^{2} )}{e ^{x} - 1})

f (x) = \frac{x + 1}{x ^{2} - 9 x}

lo g_{5} (4 - 2 x)

f (x) = \frac{( x + 1 )}{( 3 x - 1 )}

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