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

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

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domain of sqrt(16-x^2)

domain

16 - x^{2}

- 4 \leq x \leq 4

Interval Notation

[- 4, 4]

Solution steps

Find non-negative values for radicals:

- 4 \leq x \leq 4

The function domain

- 4 \leq x \leq 4

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

\frac{2}{3}

f (x) = x^{3} + 4 x^{2}

\frac{x ^{2} + 8}{x ^{2} - 4}

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