What is the domain of f(x)=2+sqrt(9-x^2) ?

The domain of f(x)=2+sqrt(9-x^2) is -3<= x<= 3

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domain of f(x)=2+sqrt(9-x^2)

domain

f (x) = 2 + 9 - x^{2}

- 3 \leq x \leq 3

Interval Notation

[- 3, 3]

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

The function domain

- 3 \leq x \leq 3

\frac{3}{x + 2} + 1

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

- x + 49

\frac{6 x - 5}{3}

2 x^{2} + 7 x

What is the domain of f(x)=2+sqrt(9-x^2) ?
The domain of f(x)=2+sqrt(9-x^2) is -3<= x<= 3