What is the domain of y= 1/(sqrt(x-9)) ?

The domain of y= 1/(sqrt(x-9)) is x>9

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domain of y= 1/(sqrt(x-9))

domain

y = \frac{1}{x - 9}

x > 9

Interval Notation

(9, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 9

Find undefined (singularity) points:

x = 9

Combine real regions and undefined points for final function domain

x > 9

3 - 2 x^{2} - x

\frac{ln ( 5 x )}{x}

9 - x + 4

2 sin (x) - x

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