# Solving $\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$

How do I solve:

$$\frac{1}{x^2-1} \geqslant \frac{1}{x+1}$$

Here's my attempt to solve it: I brought the fraction on the right side to the left side, then I've found the common denominator, $$(x-1)(x+1)$$, and after all the computations, this is my situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$ Now, I've put the numerator and the denominator in a system of inequalities: (I don't know how to type the system of inequalities in mathjax, sorry)

$$x<2$$ $$x < -1 \space \text{or} \space x>1$$ My result is (I found the intersection): $$-12$$

Let me know if the result is wrong. Additionally, because in the test I'll have only 50 minutes (lots of questions), this method requires a lot of time; if you know a quicker method, let me know.

EDIT (solved in a quicker way):

1. when you are in this situation: $$\frac{2-x}{(x-1)(x+1)} \geqslant 0$$ change the sign of the inequality and the signs that appear in the numerator,
2. then, check each interval in the normal way, but don't forget you have changed the sign; therefore, if you had a greater sign, and you changed it, you have to take only negative intervals (not positive intervals, in order to express the correct solution).
• Essentially you need to have it that both the numerator and denominator have the same sign - so either $x <2, |x| > 1$ or $x > 2, |x| < 1$ – Dhanvi Sreenivasan Dec 8 '20 at 6:43
• Another way could be using another system of inequalities. In the first one, I'd use the greater sign and in the other one I'd use the less than sign, in order to avoid the absolute value. Is that correct? – Gabriel Burzacchini Dec 8 '20 at 6:49
• I'm afraid of doing more complicated and long stuff with the absolute value (I would have to split the exercise in more possible cases). – Gabriel Burzacchini Dec 8 '20 at 6:51
• A system is typeset with $\text{\begin{cases}...\\\\...\\\\...\\end{cases}}$ where the ellipses stand for any expression. – Yves Daoust Dec 8 '20 at 7:48

We note on the side that $$x^2\ne1$$ and multiply by $$x^2-1$$, so need to distinguish two cases:

$$\begin{cases}x^2-1>0\to 1\ge x-1,\\x^2-1<0\to 1\le x-1.\end{cases}$$

The second case is impossible and we are left with

$$x<-1\lor1

You can also work in systematic way, using a table of sign variations,

$$\begin{array}{}&&-1&&1&&2\\\hline x+1&-&0&+&+&+&+&+\\x-1&-&-&-&0&+&+&+\\2-x&+&+&+&+&+&0&-\\\hline&+&|&-&|&+&0&-\end{array}$$

Use the difference of two squares to get $$\frac{1}{(x+1)(x-1)} ≥ \frac{1}{x+1}$$.

Case 1: If $$x + 1 > 0 \Rightarrow x > -1$$, we have:

$$\frac{1}{x-1} ≥ 1 \tag{x \ne -1, 1}$$

The $$x \ne -1, 1$$ comes from the denominator $$(x+1)(x-1)$$, which is undefined when $$x = -1, 1$$.

If $$x - 1 > 0 \Rightarrow x > 1$$, then multiplying both sides by $$x-1$$ gives $$1 ≥ x - 1 \Rightarrow x ≤ 2$$. If $$x - 1 < 0$$, then because $$x > -1$$, there are no solutions for this case. Hence the intersection of $$x > -1, x > 1$$, and $$x ≤ 2$$ imply $$1 < x ≤ 2$$.

Case 2: If $$x + 1 < 0 \Rightarrow x < -1$$, we have:

$$\frac{1}{x-1} ≤ 1 \tag{x \ne -1, 1}$$

$$x - 1 > 0$$ is not possible in this case because $$x < -1$$. Thus $$x -1 < 0$$, or just $$x < -1$$, and there are no more conditions.

So the solution to the inequality is $$x < -1$$, $$1 < x ≤ 2$$ for $$x \in \mathbb R$$.

• Of course you can do this by sketching the graph, but I wonder if this can be done in an exam. If the question is multiple choice, use whatever method you prefer. – Toby Mak Dec 8 '20 at 7:19

First write the function such that both numerator an denominator have same sign. $$\frac{x-2}{(x-1)(x+1)} \leqslant 0$$ We use the sign scheme method for this. Now first we will find the critical points. Critical points are those points where the function changes its behaviour. Usually for this problem the zeroes and poles are those points.

Critical points are $$x = 2,1,-1$$. There are three critical points with each critical point having odd frequency i.e. all critical points occur odd number of times.

Now plot the points on a number line.

1. Find the maximum critical point. In this case it's $$x =2$$. The function is positive always when the value of x is greater than the maximum critical point. So the function is positive for $$x \geq 2$$.
2. The function between $$x=2$$ and $$x = 1$$ will be negative.
3. The function between $$x=1$$ and $$x=-1$$ will be positive.
4. The function beyond $$x=-1$$ will be negative.

Tricks to solve it quickly:

1. Always write the function in the above format. So that numerator and denominator have same sign.
2. Find the critical points. Any value of $$x$$ beyond the maximum critical point will make the function positive.
3. Now as we go backwards, see how many times the critical points occurred. Since $$x=2$$ occurred once, sign of the function will change once from positive to negative. And the function will behave such that its negative till the next critical point is encountered.
4. Now once you reach the critical point $$x=1$$ you will see it occurred odd number of times again the sign transition will happen once. So the function will behave such that its positive till it encounters the next critical point.
5. Once you hit the last critical point, see the frequency since its odd, one sign change. So beyond $$x=-1$$ the function will be negative.
6. Remember poles wont be included in the final solution as you will make the function infinitely large. Also zeroes will be included since the inequality also has an equal sign.
7. Now compile the solution from these regions: you want all the negative regions, the final solution being: $$x\in (-\infty,-1) \cup (1,2]$$

I will give a completely different example for the same thing so that concepts are clear. Lets say your function was: $$\frac{(x-2)(x-3)^2(x+5)^3}{(x-1)^5x(x+1)^6} \lt 0$$

1. Format is correct, numerator denominator has same signs.
2. Find critical points with frequency (number of times each point occurred). So the critical points and frequency are : $$x = 2(1),3(2),-5(3),0(1),-1(6),1(5)$$.
3. The region greater than the maximum critical point is always positive. So for values $$x \gt 3$$ the function is always positive. Note I have not done $$x \geq 3$$ because thats not the requirement of the inequality. It will follow for every such critical point.
4. As we move backwards, observe the number of occurrences of the critical point (odd or even) and apply that many sign transitions. For example, as we reach $$x=3$$ we see it occurred two times so sign transition will be 2 times from + to - (1 transition) and from - to + (2nd transition). So overall when the values are just less than 3 the sign of function didnot change.
5. Go backwards, as you encounter critical points make sign transitions equal to the number of occurrences of the critical point. So if 2 occured once only one sign transition will happen as the function goes beyond it.

Even frequency of critical point: No sign change will happen

Odd frequency of critical point: Sign change will happen

1. Now compile the solution. In the inequality we need the function to be negative. So just compile the negative parts from the figure. The solution will be : $$x \in (-5,-1) \cup (-1,0) \cup (1,2)$$ I had to break at -1 since the function will go to $$-\infty$$. If you want the solution such that the function can go to $$-\infty$$ you can include that. Hope this helps. With practice you will solve these lenghty questions in seconds. Tested :D