What is the domain of sqrt(4-x)+sqrt(x^2-9) ?

The domain of sqrt(4-x)+sqrt(x^2-9) is x<=-3\lor 3<= x<= 4

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domain of sqrt(4-x)+sqrt(x^2-9)

domain

4 - x + x^{2} - 9

x \leq - 3 or 3 \leq x \leq 4

Interval Notation

(- \infty, - 3] \cup [3, 4]

Solution steps

Find non-negative values for radicals:

x \leq - 3 or 3 \leq x \leq 4

The function domain

x \leq - 3 or 3 \leq x \leq 4

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

- \frac{4}{x}

f (x) = \frac{3 x ^{2} + 3 x}{x ^{2} - x}

\frac{x ^{2} - 4}{x - 2}

f (x) = (4 - 3 x)^{\frac{7}{2}}

What is the domain of sqrt(4-x)+sqrt(x^2-9) ?
The domain of sqrt(4-x)+sqrt(x^2-9) is x<=-3\lor 3<= x<= 4