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

The domain of y= 1/(sqrt(9-x^2)) is -3<x<3

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

domain

y = \frac{1}{9 - x ^{2}}

- 3 < x < 3

Interval Notation

(- 3, 3)

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

- 3 < x < 3

f (x) = \frac{4}{3 - x}

- x^{2} + 1

f (x) = 1 - 7^{x}

f (x) = \frac{9 - x}{x - 3}

y = \frac{3 x + 4}{x ^{2} + 7 x + 12}

What is the domain of y= 1/(sqrt(9-x^2)) ?
The domain of y= 1/(sqrt(9-x^2)) is -3<x<3