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

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

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

domain

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

- 2 \leq x \leq 2

Interval Notation

[- 2, 2]

Solution steps

Find non-negative values for radicals:

- 2 \leq x \leq 2

The function domain

- 2 \leq x \leq 2

f (x) = - 9 x + 36

h (x) = ln (e^{x} sin (x))

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

f (x) = x^{2} - ∣ x ∣

f (x) = tan (x) cot (x) + sin (x)

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