Thread: Double Inequalities

Need help to solve this:

0 < √1-x^2 < 1

answer is -1< x < 1

How does that work?

Thanks.

Originally Posted by anonymous_maths

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

.

Squaring everything maintains the inequality...











Now look at each inequality separately.





.


But since

.


Putting it together, that means

.

Originally Posted by Prove It

But since

.

That is false. I.e., . yet .


All you have to do from here is square root everything

yes, but that's different to the answer supplied of -1< x < 1

Originally Posted by Haven

That is false. I.e., . yet .


All you have to do from here is square root everything

Correct in finding the mistake in my logic.

However, since all square numbers are nonnegative, that means if



Then .


So NOW putting the inequalities together you find that

.

Originally Posted by anonymous_maths

Need help to solve this:

0 < √1-x^2 < 1

answer is -1< x < 1

How does that work?

Thanks.

The crucial point with double inequalities is that you must have and .

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 , .

If then , x= 0.

The whole point of that is that the three points, -1, 0, and 1 separate intervals where 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 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 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 .

it's true as long as

as for (1) fix and by squaring we get which holds for any thus the solution set for (1) is again and the final solution set it's just