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

The domain of (sqrt(4-x))/(sqrt(x^2-9)) is x<-3\lor 3<x<= 4

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

domain

\frac{4 - x}{x ^{2} - 9}

x < - 3 or 3 < x \leq 4

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 3 or 3 \leq x \leq 4

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or 3 < x \leq 4

2 x^{2} - 16 x + 30

f (x) = \frac{- 4}{x}

f (x) = 2 \cdot x^{3} + 5 \cdot x^{2} - 10 \cdot x + 1

f (x) = - x^{2} + 9 x + 10

4 y - 1 = - 3 (4 - 2 x)

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