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

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

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

domain

(\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

f (x) = \frac{x ^{2} + 3 x - 2}{x ^{2} - 5 x + 6}

f (x) = \frac{( 3 x )}{7 x + 14}

f (x) = 2 x^{3} - 9 x^{2} - 24 x + 20

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

9 x - 16

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