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

The domain of f(x)=(sqrt(x-1))/(x^2-9) is 1 3

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

domain

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

1 \leq x < 3 or x > 3

Interval Notation

[1, 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 1

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

1 \leq x < 3 or x > 3

(x^{2})^{2}

\frac{2 x + 8}{- 8 x + 4}

f (x) = (x + 3)^{2} - 5

f (x) = x sin (\frac{1}{x})

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

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