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

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

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

domain

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

x < 0 or x > 9

Interval Notation

(- \infty, 0) \cup (9, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 9

Find undefined (singularity) points:

x = 9, x = 0

Combine real regions and undefined points for final function domain

x < 0 or x > 9

- 3 (x - 3)^{2} (x^{2} - 1)

f (x) = \frac{1}{x ^{2} - 10 x + 25}

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

f (x) = (x - 7)^{2}

y = - x^{2} + 7 x + 1

What is the domain of f(x)= 1/(sqrt(x^2-9x)) ?
The domain of f(x)= 1/(sqrt(x^2-9x)) is x<0\lor x>9