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

The domain of f(x)=(1/(sqrt(x)))^2-9 is x>0

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

domain

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

x > 0

Interval Notation

(0, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

x > 0

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