What is the domain of f(x)= 5/(x-4)+sqrt(x^2-9) ?

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

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domain of f(x)= 5/(x-4)+sqrt(x^2-9)

domain

f (x) = \frac{5}{x - 4} + x^{2} - 9

x \leq - 3 or 3 \leq x < 4 or x > 4

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 3

Find undefined (singularity) points:

x = 4

Combine real regions and undefined points for final function domain

x \leq - 3 or 3 \leq x < 4 or x > 4

f (x) = x^{2} - 4, x \leq - 2

\frac{4 - s ^{2}}{2 s ^{2} - 7 s - 4}

y = \frac{x}{2 ( 2 + x )}

f (t) = sin (t) + 4 cos (t) + t

\frac{3 x + 2}{x - 4}

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