Respuesta :
Since x can't be zero (it sits at the denominator of the left hand side), let's split cases:
If x is positive
It this case, we can multiply both sides by x without changing the inequality sign:
[tex] \dfrac{1}{x} > x \iff 1 > x^2 \iff x \in (-1,1) [/tex]
But since we're assuming that x is positive, we can only accept the numbers between 0 and 1.
If x is negative
It this case, in order to multiply both sides by x we have to change the inequality sign:
[tex] \dfrac{1}{x} > x \iff 1 < x^2 \iff x \in (-\infty,-1) \cup (1,\infty) [/tex]
But since we're assuming that x is negative, we can only accept the numbers which are smaller than -1.
So, putting all the pieces together, we have that the solution is given by the following interval:
[tex] (-\infty, -1) \cup (0,1) [/tex]
