1-\frac{2}{x-2}\ge 3
Values of x.
If I do it algebraically, I get x<=1 but from the graph it is between x= 1 and x=2 (asymptote)
How do I reconcile that? Am I doing something wrong here?
Once again a subtle point here, one worth thinking through for sure.
First, so everyone can see the issue:
The green shaded area shows y \ge 3, so you can see the values of x that satisfy the given equation are indeed 1 \le x \lt 2.
One clue is that, as you know, a hyperbola has a distinct break where the denominator equals 0, and the function behaves differently on the two sides of that line so we should be paying special attention there in anything we do. Furthermore, the fact that we see, as you recognized, that the actual answer is 1 \le x \lt 2 suggests that there’s a domain restriction we can focus on, which is (not coincidentally) exactly where the function’s behavior changes.
With that in mind, there are two equivalent ways to see how a crucial algebraic subtlety arises at exactly that x-value. Let’s start from the point
-\frac{2}{x-2} \ge 2
• One way to see the issue is to remember that when you take the reciprocal of both sides of an inequality, if the quantities on the left and on the right are both positive, or are both negative, then the inequality flips when you take the reciprocal. For instance, if we start with 3 \gt 2, then the reciprocals give \frac{1}{3} \lt \frac{1}{2}.
By contrast, if one side is positive while the other is negative, the inequality does not flip. For instance, 3 \gt -2 becomes \frac{1}{3} \gt -\frac{1}{2}.
• The other way to view the issue is that if you imagine cross-multiplying the (x-2) in the equation above, if (x-2) \gt 0 the value is positive and so the inequality does not flip, but if if (x-2) \lt 0 then the value is negative and so the inequality does flip.
Either way you view it, you can see that we need to treat TWO SEPARATE CASES, one for x \lt 2 and the other for x \gt 2. The result you found, x \le 1, came from the case where x \lt 2, and so automatically the entire result of that work is 1 \le x \lt 2, matching what the graph shows.
And if you work through the other case, you’ll see why it produces a null result.
I hope that all makes sense, and trust you’ll write back if not. For now, really appreciate that you’re working to make sense of all of these pieces for yourself! 
Got it. Thank you so much. Your messages have been very helpful.
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