What is the domain of y=(x+3)/(4-sqrt(x^2-9)) ?

The domain of y=(x+3)/(4-sqrt(x^2-9)) is x 5

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domain of y=(x+3)/(4-sqrt(x^2-9))

domain

y = \frac{x + 3}{4 - x ^{2} - 9}

x < - 5 or - 5 < x \leq - 3 or 3 \leq x < 5 or x > 5

Interval Notation

(- \infty, - 5) \cup (- 5, - 3] \cup [3, 5) \cup (5, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = 5, x = - 5

Combine real regions and undefined points for final function domain

x < - 5 or - 5 < x \leq - 3 or 3 \leq x < 5 or x > 5

f (x) = - x^{2} + 8 x - 13

f (x) = \frac{1}{x - 2} + 3 x

f (x) = - lo g_{\frac{1}{5}} (x + 6) - 5

y = x^{2} + 2 x - 3

f (x) = - 0. 6^{(x + 2)} + 3

What is the domain of y=(x+3)/(4-sqrt(x^2-9)) ?
The domain of y=(x+3)/(4-sqrt(x^2-9)) is x<-5\lor-5<x<=-3\lor 3<= x<5\lor x>5