Find the values for which $\frac{x}{1-x^2} > 0$ It's been quite long since high school and I forgot some stuff.
I'm trying to find the values for which $\frac{x}{1-x^2} > 0$.
Now, obviously this occurs when $x < -1$ and $x>0$, yet if you just solve the inequality systematically, you would just get $x > 0$ which is not completely true:
$$\frac{x}{1-x^2} > 0$$
$$(1-x^2)\frac{x}{1-x^2} > 0(1-x^2)$$
$$x > 0$$
Would appreciate someone explaining why this doesn't seem to work, thanks in advance.
 A: Two cases:

*

*$x>0$ and $1-x^2>0$, which means $x>0$ and $x^2<1$, i.e.,$0<x<1$.


*$x<0$ and $1-x^2<0$, which means $x<0$ and $x^2>1$, i.e., $x<-1$.
Thus the domain is $(-\infty,-1)\cup (0,1)$.
A: The problem is that when you multiply by $(1-x^2)$, you assume this expression is positive. When you multiply both sides of an inequality by a negative number the inequality sign should be reversed.
A: For rational functions (ratios of polynomials), sign changes can only occur where a factor of the numerator or denominator is zero. At such a point, the function itself will be either undefined (if the denominator vanishes) or zero (if the numerator vanishes).
So the best approach is to make a sign chart for each factor. For each factor draw a number line, label the point where the factor vanishes, and label the sign on each side of the factor.
Then combine the sign charts of all the factors multiplicatively to arrive at a sign chart for your function, taking care to mark the points with "$0$" where the numerator factors vanish and as "$*$" where the denominator factors vanish.
Finally, look at the inequality you need to solve. Since it is of the form "$f(x)>0$", you will include only the parts of the number line where the overall sign is positive (not zero!). If instead it were "$f(x)\geq 0$", you would also have to include the points where the overall sign equal zero (typically transition points where the numerator factors vanish), and so forth.
Example:
Solve $\dfrac{x+1}{x-2}\leq 0$.
Solution: The factors are $(x+1)$ and $(x-2)$, which vanish at $-1$ and $2$ respectively. Sign charts:
number line:    _________-1______________2___________
sign of (x+1):  ----------0++++++++++++++++++++++++++
sign of (x-2):  -------------------------0+++++++++++
combined:       ++++++++++0--------------*+++++++++++
                          ^              ^
                         zero         undefined

The combined sign is determined by first marking points where the function is zero "$0$" or undefined "$*$"; the sign on the interval in between such points is constant, and will be negative if there is an odd number of negative factors in the charts above it, and positive otherwise.
Note that after combining factors by division, the function is zero at $-1$ and undefined at $2$.
For our solution we want points on the number line where the function is negative or zero (since we are solving an inequality of the form $f(x)\leq 0$).
The solution is $[-1,2)$, which includes the left endpoint but not the right endpoint.
If the inequality were $f(x)>0$ instead, we would only want points where the combined sign is positive (not zero), so the solution would be $(-\infty,-1)\cup (2,\infty)$.
Now you should be able to do something similar. You will have three sign charts to combine.
(BTW, your solution doesn’t work because you have ignored the sign contribution of the denominator.)
A: You can multiply both sides of the inequality (and keep it the same) if the function you're multiplying with has the ame sign as the inequality sign.
For example, you can safely multiply (x^2+1), because it is positive everywhere.
But is (1-x^2) positive everywhere?
A: You need the intersection of sets $\{x| x>0\} $ and $\{x| (1-x^2)>0\}$ which happens for $0<x<1$.
Another possibility is to get the intersection of $\{x| x<0\} $ and $\{x| (1-x^2)<0\}$ which happens for $x<-1$.
A: The problem is, you seem to believe that you can multiply an inequality by any number, just like in equations. However, note that $5>3$ is correct. Multiply by $-1$ to get $-5<-3$, which is obviously false. You must note here, that

An inequality can be multiplied by any positive number, and will still hold true. But multiplying by a negative number, we need to reverse the sign.

This is why, in inequalities, often to get rid of the denominator, we multiply by the square of the denominator (it is always positive), and not the denominator. So, for your case,
$$x(1-x^2)>(1-x^2)^2\implies x-x^3>1-2x^2+x^4$$
which is not very nice. Instead of this, you can use the simple fact that
$$\frac ab > 0 \iff a,b>0 \text{ or }a,b<0$$
And divide the problem in two cases and solve separately.
Hope this helps. Ask anything if not clear :)
A: Here's an easy way to handle this:
You have $\frac {x} {1-x^2} > 0 \implies \frac {x} {(1+x)(1-x)} > 0$
Multiply both sides by $(1+x)^2(1-x)^2$. You don't have to worry about the direction of the inequality as that expression is non-negative for real $x$.
So, after cancellation, you get $x(1+x)(1-x) >0$
It's now just a matter of sketching an already fully factorised cubic (so you know the $x$ intercepts). Note that the lead coefficient will be negative so you know $y \to \pm \infty$ respectively as $x \to \mp \infty$. That allows you to complete the sketch, which immediately gives you the solution set $(x<-1) \cup (0<x<1)$.
Your mistake was in multiplying by an expression $(1-x^2)$ whose sign you cannot assume for all reals. That problem does not exist in my solution.
