Need help to solve this:
0 < √1-x^2 < 1
answer is -1< x < 1
How does that work?
Thanks.
I take it that this is
$\displaystyle 0 \leq \sqrt{1 - x^2} \leq 1$.
Squaring everything maintains the inequality...
$\displaystyle 0 \leq 1 - x^2 \leq 1$
$\displaystyle -1 \leq -x^2 \leq 0$
$\displaystyle 1 \geq x^2 \geq 0$
$\displaystyle 0 \leq x^2 \leq 1$
Now look at each inequality separately.
$\displaystyle x^2 \leq 1$
$\displaystyle |x| \leq 1$
$\displaystyle -1 \leq x \leq 1$.
But since $\displaystyle 0\leq x^2$
$\displaystyle 0 \leq x$.
Putting it together, that means
$\displaystyle 0 \leq x \leq 1$.
Correct in finding the mistake in my logic.
However, since all square numbers are nonnegative, that means if
$\displaystyle 0 \leq x^2$
Then $\displaystyle -\infty \leq x \leq \infty$.
So NOW putting the inequalities together you find that
$\displaystyle -1 \leq x \leq 1$.
The crucial point with double inequalities is that you must have $\displaystyle 0\le \sqrt{1- x^2}$ and $\displaystyle \sqrt{1- x^2}\ge 1$.
Personally, I think the simplest way to solve complicated inequalities is to solve the associated equation first.
If [/tex]0= \sqrt{1- x^2}[/tex] then $\displaystyle 0= 1- x^2$, $\displaystyle x= \pm 1$.
If $\displaystyle \sqrt{1- x^2}= 1$ then $\displaystyle 1- x^2= 1$, x= 0.
The whole point of that is that the three points, -1, 0, and 1 separate intervals where $\displaystyle \sqrt{1- x^2}$ is less than 0, greater than 0, or does not exist. Try one value of x in each of the intervals x< -1, -1< x< 0, 0< x< 1, and x> 1.
For example, if x= -2< -1, then $\displaystyle \sqrt{1- x^2}= \sqrt{-3}$ which is not a real number so does not satisfy the inequality. No value of x less than -1 satisfies the inequality.
If x= -1/2, between -1 and 0, then [tex]\sqrt{1- x^2}= \sqrt{1- 1/4}= \sqrt{3}{2}. which is positive and less than 1. Every value of x between -1 and 0 satisfies the inequality.
If x= 1/2, between 0 and 1, then [tex]\sqrt{1- x^2}= \sqrt{1- 1/4}= \sqrt{3}{2}. which is positive and less than 1. Every value of x between 0 and 1 satisfies the inequality.
Finally, if x= 2, between 0 and 1, then $\displaystyle \sqrt{1- x^2}= \sqrt{-3}$ which is not a real number so does not satisfy the inequality. No value of x larger than 1 satisfies the inequality.
Since the numbers -1, 0, and 1 make the two sides equal, the solution set is $\displaystyle -1\le x\le 1$.
$\displaystyle \sqrt{1-x^2}\ge0$ it's true as long as $\displaystyle |x|\le1.$
as for $\displaystyle \sqrt{1-x^2}\le1,$ (1) fix $\displaystyle |x|\le1$ and by squaring we get $\displaystyle x^2\ge0$ which holds for any $\displaystyle x,$ thus the solution set for (1) is again $\displaystyle |x|\le1,$ and the final solution set it's just $\displaystyle |x|\le1.$