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Give the interval(s) on which the function is continuous.
g(t) = 1/√16-t^2

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The function g(t) is defined as:

g(t) = 1/√(16-t^2)

The function is continuous for all values of t that satisfy the following conditions:

The denominator is non-zero:

The denominator of the function is √(16-t^2). Therefore, the function is undefined when 16-t^2 < 0, or when t is outside the interval [-4,4].

There are no vertical asymptotes:

The function does not have any vertical asymptotes, because the denominator is always positive.

Thus, the function g(t) is continuous on the interval [-4,4].

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