What is the domain of f(x)= 1/(x-5)+sqrt(x^2-9) ?

The domain of f(x)= 1/(x-5)+sqrt(x^2-9) is x 5

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domain of f(x)= 1/(x-5)+sqrt(x^2-9)

domain

f (x) = \frac{1}{x - 5} + x^{2} - 9

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

Interval Notation

(- \infty, - 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

Combine real regions and undefined points for final function domain

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

+ 5 x + 35

\frac{2}{2 + 2 ^{\frac{1}{x}}}

x + \frac{9}{x}

\frac{4}{x} - x + 5

(- 2.1) (4.3)

What is the domain of f(x)= 1/(x-5)+sqrt(x^2-9) ?
The domain of f(x)= 1/(x-5)+sqrt(x^2-9) is x<=-3\lor 3<= x<5\lor x>5