i have no problem solving inequalities but this one has me stumped, one over x is greater than x. For what real number is 1/x > x ?
Do some manipulation:
1/x > x (multiply both sides by 1/x)
1 > x^2 (take the square root of both sides)
+, - 1 = x
Check intervals:
- infinity to - 1
(-2)
1/-2 > -2 YES
-1 to 1
1/2 works, but -1/2 doesn't...hmmm
1/(1/2) > 1/2
2 > 1/2 YES
1/(-1/2) > -1/2
-2 > -1/2 NO
1 to infinity
(2)
1/2 > 2 NO
Solution
X < -1
0 < x < 1
A very general way of solving complicated inequalities: solve the equation first.
To solve , first look at . That's easy: multiplying both sides by x, which has roots . The point of that is that functions can change form ">" to "<" only where they are "=" or where they are not continuous. The functions involved here are equal at x= 1 or x= -1 and not continuous at x= 0 where the left side is not defined. Now we need only check on point in each of the intervals , , , and .
For example:
-2< -1 and if x= -2, .
and if , .
and if ,
1< 2 and if x= 2,
That is, for x< -1 or 0< x< 1.
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