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

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

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

domain

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

- 3 < t < 3

Solution steps

Find non-negative values for radicals:

- 3 \leq t \leq 3

Find undefined (singularity) points:

t = 3, t = - 3

Combine real regions and undefined points for final function domain

- 3 < t < 3

(150, 147.9), (165, 163.4)

f (x) = - \frac{194}{63} x + \frac{6133}{6300}

7 - x

f (x) = 64 x^{2}

(3, - 5), (- 4, 2)

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