What is the domain of q(x)=(x-3)/(x^2+x-12) ?

The domain of q(x)=(x-3)/(x^2+x-12) is x 3

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domain of q(x)=(x-3)/(x^2+x-12)

domain

q (x) = \frac{x - 3}{x ^{2} + x - 12}

x < - 4 or - 4 < x < 3 or x > 3

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 3, x = - 4

The function domain

x < - 4 or - 4 < x < 3 or x > 3

\frac{x ^{2} - 4 x + 4}{x ^{4} - 8 x ^{2} + 16}

f (x) = 10 - 9 x

f (x) = - \frac{12}{2 x - 9}

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

g (x) = \frac{x ^{2}}{( x - 3 ) ( x + 4 ) ( x + 1 )}

What is the domain of q(x)=(x-3)/(x^2+x-12) ?
The domain of q(x)=(x-3)/(x^2+x-12) is x<-4\lor-4<x<3\lor x>3