What is the domain of sqrt(16-x^2)-sqrt(x^2-1) ?

The domain of sqrt(16-x^2)-sqrt(x^2-1) is -4<= x<=-1\lor 1<= x<= 4

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domain of sqrt(16-x^2)-sqrt(x^2-1)

domain

16 - x^{2} - x^{2} - 1

- 4 \leq x \leq - 1 or 1 \leq x \leq 4

Interval Notation

[- 4, - 1] \cup [1, 4]

Solution steps

Find non-negative values for radicals:

- 4 \leq x \leq - 1 or 1 \leq x \leq 4

The function domain

- 4 \leq x \leq - 1 or 1 \leq x \leq 4

y = \frac{- 4 - 3 x}{x - 5}

y = \frac{ln ∣ x ∣}{( x - 3 )}

\frac{2 x}{3 x - 3}

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

What is the domain of sqrt(16-x^2)-sqrt(x^2-1) ?
The domain of sqrt(16-x^2)-sqrt(x^2-1) is -4<= x<=-1\lor 1<= x<= 4