What is the domain of f(x)= 1/(sqrt(4-t^2)) ?

The domain of f(x)= 1/(sqrt(4-t^2)) is -2<t<2

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domain of f(x)= 1/(sqrt(4-t^2))

domain

f (x) = \frac{1}{4 - t ^{2}}

- 2 < t < 2

Interval Notation

(- 2, 2)

Solution steps

Find non-negative values for radicals:

- 2 \leq t \leq 2

Find undefined (singularity) points:

t = 2, t = - 2

Combine real regions and undefined points for final function domain

- 2 < t < 2

f (x) = ln (2 - x) + 1

f (x) = \frac{( x + 6 ) ^{2}}{8 x - 7}

f (x) = - sin (x)

\frac{x ^{2} - 16}{x ^{2} - 81}

f (x) = 9 cot (2 x - \frac{π}{2}) - 3

What is the domain of f(x)= 1/(sqrt(4-t^2)) ?
The domain of f(x)= 1/(sqrt(4-t^2)) is -2<t<2