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 $\displaystyle \frac{1}{x}> x$, first look at $\displaystyle \frac{1}{x}= x$. That's easy: multiplying both sides by x, $\displaystyle \frac{1}= x^2$ which has roots $\displaystyle x= \pm 1$. 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 $\displaystyle (-\inf, -1)$, $\displaystyle (-1, 0)$, $\displaystyle (0, 1)$, and $\displaystyle (1, \infty)$.
For example:
-2< -1 and if x= -2, $\displaystyle \frac{1}{x}= -\frac{1}{2}> -2= x$.
$\displaystyle -2< -\frac{1}{2}< 0$ and if $\displaystyle x= -\frac{1}{2}$, $\displaystyle \frac{1}{x}= -2< -\frac{1}{2}= x$.
$\displaystyle 0< \frac{1}{2}< 1$ and if $\displaystyle x= \frac{1}{2}$, $\displaystyle \frac{1}{x}= 2> \frac{1}{2}= x$
1< 2 and if x= 2, $\displaystyle \frac{1}{x}= \frac{1}{2}< 2= x$
That is, $\displaystyle \frac{1}{x}> x$ for x< -1 or 0< x< 1.
This same question was asked here.
http://www.mathhelpforum.com/math-he...questions.html
We are told:
$\displaystyle \frac{1}{x} > x$.
First, we have to note that $\displaystyle x \neq 0$.
If we are trying to solve this inequality, we need to take into account two possibilities: $\displaystyle x > 0$ and $\displaystyle x < 0$.
Case 1: $\displaystyle x > 0$.
$\displaystyle \frac{1}{x} > x$
$\displaystyle 1 > x^2$
$\displaystyle x^2 < 1$
$\displaystyle |x| < 1$
$\displaystyle -1 < x < 1$.
But since $\displaystyle x > 0$
$\displaystyle 0 < x < 1$.
Case 2: $\displaystyle x < 0$
$\displaystyle \frac{1}{x} >x$
$\displaystyle 1 < x^2$
$\displaystyle x^2 > 1$
$\displaystyle |x| > 1$
$\displaystyle x < -1$ or $\displaystyle x > 1$.
But since $\displaystyle x < 0$ we have $\displaystyle x < -1$.
So putting everything together:
$\displaystyle x < -1$ or $\displaystyle 0 < x < 1$.