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

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

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

domain

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

\frac{2}{x + 3}

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

y = x^{2} - x - 6

f (x) = \frac{4 x ^{2} + 2 x - 1}{2 x ^{3}}

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 17

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