What is the domain of f(x)=(sqrt(3+x))/(9-x) ?

The domain of f(x)=(sqrt(3+x))/(9-x) is -3 9

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domain of f(x)=(sqrt(3+x))/(9-x)

domain

f (x) = \frac{3 + x}{9 - x}

- 3 \leq x < 9 or x > 9

Interval Notation

[- 3, 9) \cup (9, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 3

Find undefined (singularity) points:

x = 9

Combine real regions and undefined points for final function domain

- 3 \leq x < 9 or x > 9

f (x) = \frac{7 x}{x ^{2} - 1}

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

f (x) = 2 arcsin (\frac{x}{2}) + \frac{π}{4}

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

x^{2} + 11, x \geq 0

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