What is the domain of sqrt(16-t^2) ?

The domain of sqrt(16-t^2) is -4<= t<= 4

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domain of sqrt(16-t^2)

domain

16 - t^{2}

- 4 \leq t \leq 4

Interval Notation

[- 4, 4]

Solution steps

Find non-negative values for radicals:

- 4 \leq t \leq 4

The function domain

- 4 \leq t \leq 4

f (x) = \frac{3 x}{13 + 4 x}

f (x) = \frac{x ^{2} - 5 x + 6}{x + 4}

f (x) = \frac{1}{x ^{4} - 5 x ^{2} + 4}

\frac{1}{2 x}

f (x) = x - 4 + 1