What is the domain of f(x)=3+sqrt(16-x^2) ?

The domain of f(x)=3+sqrt(16-x^2) is -4<= x<= 4

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domain of f(x)=3+sqrt(16-x^2)

domain

f (x) = 3 + 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{1}{x})

f (x) = 1 - e^{- 0.5 x}

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

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

y = - 2 x + 2

What is the domain of f(x)=3+sqrt(16-x^2) ?
The domain of f(x)=3+sqrt(16-x^2) is -4<= x<= 4