What is the domain of y=(x+6)/(4-sqrt(x^2-9)) ?

The domain of y=(x+6)/(4-sqrt(x^2-9)) is x 5

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domain of y=(x+6)/(4-sqrt(x^2-9))

domain

y = \frac{x + 6}{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) = \frac{x ^{2} - 81}{x - 9}

y = (x + 1)

y = - lo g_{3} (x) + 2

y = \frac{sin ( 4 x )}{x}

f (x) = 3 \cdot (\frac{3}{5})^{x + 2} - 4

What is the domain of y=(x+6)/(4-sqrt(x^2-9)) ?
The domain of y=(x+6)/(4-sqrt(x^2-9)) is x<-5\lor-5<x<=-3\lor 3<= x<5\lor x>5