## anonymous one year ago 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.

1. anonymous

$\sqrt{1-x^2}$

2. Vocaloid

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

3. anonymous

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

4. Vocaloid

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

5. anonymous

Thanks

6. SolomonZelman

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

7. mathstudent55

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). |dw:1437618309443:dw|

8. mathstudent55

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. |dw:1437618438257:dw|

9. mathstudent55

Now you test each of the three regions to see which ones are part of the solution. To test each region, choose a point in teat region and test is. If the point works, then the entire region works. Let's test -2 $$1 - (-2)^2 \ge 0$$ $$1 - 4 \ge 0$$ $$-3 \ge 0$$ is false, so left of -1 is not part of the solution. Now let's test 0 $$1 - 0^2 \ge 0$$ $$1 - 0 \ge 0$$ $$1 \ge 0$$ is true, so between -1 and 1 is part of the solution. Now let's test 2 $$1 - ^2 \ge 0$$ $$1 - 4 \ge 0$$ $$-3 \ge 0$$ is false, so to the right of 1 is not part of the solution. |dw:1437618822904:dw|