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

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

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

domain

3 - x + 16 - x^{2}

- 4 \leq x \leq 3

Interval Notation

[- 4, 3]

Solution steps

Find non-negative values for radicals:

- 4 \leq x \leq 3

The function domain

- 4 \leq x \leq 3

f (x) = \frac{4 x}{- 3 x + 1}

f (x) = 2 x - 5 + 7

f (x) = cos (x - \frac{π}{2})

\frac{- 4 - 6 x}{9 x - 5}

g (x) = 2 x - 5

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