An inequality about 1/x looks like a question about small positive numbers, so nobody checks x = −1, and the answer that loses half the number line goes unnoticed. The fix is a method that cannot skip a region.
Put the slider on −1 before you say anything, and ask the class whether x = −1 is a solution of 1/x < 2. The two middle readouts answer it for them: the inequality says yes and the naive answer says no.
The two bands under the axis are the argument. The true band has a piece running off the left of the screen and the naive band does not, and the label says so. Do not explain the band; ask what is missing from the lower one.
Then drag the slider to 0 deliberately. The readout says undefined and the verdict says x = 0 is excluded. That is worth doing, because the zero of the denominator being a critical value is the second half of this sub-topic's difficulty and students treat it as a technicality rather than a boundary.
| Question | Answer |
|---|---|
| 1. The positive boundary | 0.5, where 1/x = 2. |
| 2. 1/0.4 | 2.5, which is above 2. |
| 3. How many regions | 3, from the two critical values 0 and 0.5. |
| 4. x³ − x at −0.5 | 0.375, positive. |
| 5. Why x > 0.5 is incomplete | B. The negative case reverses the inequality. |
Question 3 is the diagnostic one. A student who answers 2 has found the boundary from the numerator and not the one from the denominator, and that is the error that produces a single tidy interval on every rational inequality they will meet.
1 markThe critical values, including the zero of any denominator.
1 markTesting each region, or a sign diagram.
1 markThe answer, with "or" between the pieces when there are two.
The middle mark has to be visible. Three substitutions written out, or a sign diagram drawn, earns it. A correct answer with no working does not, and students find that genuinely unfair until it is explained that the paper is marking the method.
| They wrote | What happened |
|---|---|
| x > 0.5 as the whole answer | The error this page exists for. Multiplied through by x. Put the slider on −1 instead of explaining. |
| 0 < x < 0.5 | Got the direction backwards. They have found the one region that is NOT in the solution, which usually means they tested one value and trusted it. |
| x < 0.5 | Split into cases and then took the union carelessly, keeping x < 0.5 from the negative case without intersecting it with x < 0. The intersection step is the one that gets dropped. |
| 2 on question 3 | One critical value instead of two. The denominator's zero is missing. This is the single most useful thing to catch on this page. |
| 0.4 on question 2 | Gave x rather than 1/x. Mechanical, but check they are reading the expression and not the question's numbers. |
| 0.25 on question 2 | Computed 1/4, treating 0.4 as a quarter. Worth a word about estimating: 1/0.4 must be above 2 because 0.4 is below a half. |
| −0.375 on question 4 | Evaluated at +0.5. The sign decides whether the region is in the answer, so this error flips a whole interval. |
| −0.625 | Subtracted +0.5 instead of −0.5. Subtracting a negative, and it is worth writing the brackets in. |
| A single interval on any rational inequality | Almost always one of the two errors above. Ask how many critical values they found before looking at anything else. |
"Can I just use the graph?" On Paper 2, yes, and the guide expects it: the content row says these may be solved graphically or using technology. The graph is also the more reliable method, because the regions are visible and a missing piece is obvious. On Paper 1 there is no graph, which is the reason to make the critical-value method secure rather than treating it as the fallback.
"Why not just always split into cases?" Because the case method needs an intersection at the end of each branch and that is where it goes wrong: students get x < 0.5 from the negative case and forget to intersect it with x < 0. The critical-value method has no such step. Teach cases once so they understand why the answer is a union, then move to critical values for actually doing it.
"Do I include the boundary?" Strict inequality excludes it, non-strict includes it, and a value that makes a denominator zero is always excluded whichever the inequality. That last part catches people on questions like 1/x ≤ 2, where the answer is x < 0 or x ≥ 0.5: the 0.5 is in and the 0 is out, in the same answer.
"What if a root is repeated?" Then the sign does not change there, and the alternating pattern breaks. x(x − 1)² > 0 is positive on both sides of 1. Worth one example, because students who have learnt "the signs alternate" will apply it and lose a region. The safe statement is that the signs alternate at a simple root.
| Demonstrate | Graph both sides rather than the difference: Y1=1÷X and Y2=2. "Where is the curve below the line" is a question students can answer by looking, and "where is this expression below zero" is not. Then point at the left branch, which never crosses the line at all. |
|---|---|
| Where they stick | The default window. Starting at −10 the left branch is squashed against the axis and looks like it might cross; −3 to 3 with y from −4 to 6 shows the real behaviour. |
| The check | The Casio's inequality graph type shades the solution. The shading appears in two separate places, which is the union made visual, and it is the fastest way to convince a class that a two-part answer is normal rather than exotic. |
Worth saying out loud that the intersection finder gives 0.5 and says nothing at all about the left branch. The machine answers the question it was asked, and "where do they cross" is not the same question as "where is one below the other".
| Step | What |
|---|---|
| 1 | Ask them to solve 1/x < 2 cold, with no method given. Collect answers. |
| 2 | Most will have x > 0.5. Ask whether x = −1 works, and let them check it themselves. |
| 3 | The figure, slider at −1, then the two bands. |
| 4 | Why: multiplying by a negative reverses the inequality. Do the two cases properly on the board. |
| 5 | Then the critical-value method, as the thing to use from now on. |
| 6 | Count critical values, add one, test each region. Three substitutions. |
| 7 | x³ > x, where the critical values are just roots and there are four regions. |
| 8 | The slider to 0, and the denominator's zero as a boundary that is always excluded. |
Do not say "multiply up to clear the fraction". It is the standard instruction for equations and it is invalid for inequalities, and students do not hear the difference. If you want one sentence: you may only multiply an inequality by something whose sign you know.
Do not let an answer stand as a single interval without being challenged. Ask "how many pieces?" every time, out loud, until it is the first thing they ask themselves. A rational or cubic inequality usually has two, and the single-interval answer is the signature of both common errors.