What is the domain of f(x)=1-sqrt(16-x^2) ?

The domain of f(x)=1-sqrt(16-x^2) is -4<= x<= 4

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domain of f(x)=1-sqrt(16-x^2)

domain

f (x) = 1 - 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

sin (4 x)

f (x) = x^{2} + 4 x + 5

f (x) = \frac{2}{5 - 2 x}

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

\frac{x - 1 - 1}{x - 1 + 1}

What is the domain of f(x)=1-sqrt(16-x^2) ?
The domain of f(x)=1-sqrt(16-x^2) is -4<= x<= 4