What is the domain of $f (x) = arcsin (1 - x^{2})$ $f(x) = arcsin(1-x^2)$ ?

Question

7078 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-25T15:04:33

The domain of $f (x)$ $f(x)$ is the interval $(- \sqrt{2}, \sqrt{2})$ $(-sqrt2,sqrt2)$

The domain of arcsin is the interval $[- 1, 1]$ $[-1,1]$ , so:

$- 1 \leq (1 - x^{2}) \leq 1$ $-1 <= (1-x^2) <= 1$

$- 1 - 1 \leq (- x^{2}) \leq 1 - 1$ $-1 -1 <= (-x^2) <=1-1$

$- 2 \leq (- x^{2}) \leq 0$ $-2 <= (-x^2) <= 0$

$0 \leq x^{2} \leq 2$ $0 <= x^2 <=2$

$- \sqrt{2} \leq x \leq \sqrt{2}$ $-sqrt(2) <= x <= sqrt2$

What is the domain of f(x) = arcsin(1-x^2)f(x)=arcsin(1−x2)f(x) = arcsin(1-x^2)?