How do I algebraically find the domain of sqrt (1 - x^2)? (i'll re-post as an equation)
I know the answer is [-1,1]. But how do I go about solving it algebraically.

Question

How do I algebraically find the domain of sqrt (1 - x^2)?  (i'll re-post as an equation)
I know the answer is [-1,1]. But how do I go about solving it algebraically.

anonymous · Answer

$$\sqrt{1-x^2}$$

vocaloid · Answer

set the expression underneath the radical greater or equal to 0

$$1-x^{2} \ge 0 $$

anonymous · Answer

Is what I have below correct?

$$1 - x^2 \ge 0  $$ $$1 \ge x^2$$
$$\sqrt{1}\ge \sqrt{x^2}$$

$$\pm1\ge x$$

vocaloid · Answer

trying to come up with a good way to explain this, sorry

$$1 \ge x^{2}$$

results in

$$-1 \le x \le 1$$

anonymous · Answer

Thanks

solomonzelman · Answer

Yes, you are correct.
√1=√(x²)
1=|x|
x=±1

mathstudent55 · Answer

The radicand must be non-negative, or

$1 - x^2 \ge 0$

To solve this inequality, factor the left side and solve it as an equation:

$(1 + x)(1 - x) = 0$

$1 + x = 0$   or   $1 - x = 0$

$x = -1$   or   $x = 1$

Now you have 2 points of interest.
These two points of interest divide the number line into 3 regions (not counting the points themselves).

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mathstudent55 · Answer

Since the inequality has the symbol $\ge$ which contains =, the points of interest are part of the solution, so you put closed dots on them.

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