Multiplying by x throws half the answer away

x = −1.0

−1.0001/x

yesis 1/x < 2?

nodoes x > 0.5 say so?

they disagreeso

The two bands under the axis are the two answers. One of them is missing the entire left half of the number line.

The figure needs JavaScript. It draws y = 1/x and the line y = 2. The curve is below the line for every negative x and for x > 0.5, and above it on 0 < x < 0.5. So the solution of 1/x < 2 is x < 0 or x > 0.5. Multiplying through by x gives only x > 0.5, which omits every negative solution: at x = −1 the left side is −1 and at x = −3 it is −0.333. Everything below stands on its own.

Why multiplying through is not allowed

Multiplying an inequality by a positive number keeps its direction. Multiplying by a negative number reverses it. So multiplying by x, when you do not know the sign of x, does two different things and you cannot know which.

Doing it properly means splitting into cases:

If x > 0: multiply and keep the direction. 1 < 2x, so x > 0.5. Combined with x > 0 this gives x > 0.5.
If x < 0: multiply and reverse it. 1 > 2x, so x < 0.5. Combined with x < 0 this gives all of x < 0.

The answer is the union: x < 0 or x > 0.5. The naive method found case 1 and silently assumed case 2 away.

Note that case 2 came out as "every negative x", not as some interval. That is why the error is so large: the lost piece is not a sliver, it is half the number line.

The safer method, which never needs cases. Get everything on one side and ask where that expression is positive:

1/x − 2 < 0, so (1 − 2x)/x < 0

Now find the critical values, where the expression is zero or undefined: x = 0.5 from the top, and x = 0 from the bottom. Those two split the line into three regions, and you test one value in each:

Region	Test	1/x	Under 2?
x < 0	x = −1	−1	yes
0 < x < 0.5	x = 0.4	2.5	no
x > 0.5	x = 0.6	1.667	yes

Three substitutions and the answer is settled, with no cases, no reversing and nothing to forget. The zero of the denominator is a critical value too, which is the step people leave out.

Move the slider to −1 and read the two middle readouts. The inequality says yes and the naive answer says no. Keep going left and they disagree at every single value. Then come back past 0.5 and they agree again.

That is the shape of this error: it is correct on the half of the line people check. Nobody tests a negative value in an inequality about 1/x, because the question looks like it is about small positive numbers.

Polynomial inequalities, up to degree three

The same method, and now the critical values are just the roots. Solve x³ > x.

One side: x³ − x > 0, which factorises as x(x − 1)(x + 1) > 0.
Critical values −1, 0 and 1: four regions.
Test one value in each: at x = −2 it is −6, at x = −0.5 it is 0.375, at x = 0.5 it is −0.375, at x = 2 it is 6.
So the answer is −1 < x < 0 or x > 1.

Two pieces again, and the signs alternate between consecutive roots, which is worth noticing: once you have one region's sign the rest follow unless a root is repeated.

Never divide an inequality by x to "simplify" x³ > x into x² > 1. It is the same error as before and it loses the piece between −1 and 0.

On the GDC: reading an inequality off a graph

The guide allows technology for these, and a graph is genuinely the most reliable method: the regions are visible and cannot be forgotten, which is the whole failure mode of the algebraic route.

When you may use it. Analysis Paper 1 is non-calculator, so the critical-value method has to be secure. On Paper 2 graph both sides and read off where one is above the other, and the guide expects exactly that.

TI-Nspire CX II

A Graphs page with f1(x)=1/x and f2(x)=2. Plot both sides rather than the difference; "where is the curve below the line" is easier to read than "where is this below zero"
menu → Window / Zoom → Window Settings, x from −3 to 3 and y from −4 to 6. The default window hides how the left branch behaves
menu → Analyze Graph → Intersection finds (0.5, 2), which is the only crossing
Then look at the left branch and notice it never crosses the line at all, so the whole of it is in the solution. That is the piece the algebra loses, and on the screen it is unmissable

Casio fx-CG50

MENU → Graph with Y1=1÷X and Y2=2, then F6 DRAW
SHIFT F3 V-Window and type Xmin −3, Xmax 3, Ymin −4, Ymax 6
SHIFT F5 G-SOLVE → ISCT gives x = 0.5
The Inequality graph type is also available: F3 TYPE → F6 → Y< shades the region, which makes a union answer obvious because the shading is in two separate places

The mark people lose. Giving one interval when the answer is a union. It happens two ways: multiplying through by something of unknown sign, and forgetting that the zero of a denominator is a critical value. Both produce a tidy single interval, which is what makes them convincing. The habit: count the critical values, add one, and that is how many regions you must test. Two critical values means three regions, every time.

Your turn

1. For 1/x < 2, find the positive boundary value.

2. Evaluate 1/x at x = 0.4.

3. How many regions do the critical values of 1/x < 2 split the number line into?

4. For x³ − x, evaluate it at x = −0.5.

5. Why is x > 0.5 not the full solution of 1/x < 2?

Question 5. Why is x greater than 0.5 not the full solution of 1 over x less than 2? Because 0.5 should not be included. Because multiplying by x reverses the inequality when x is negative, and that case gives every negative x as a solution too. Because 1/x is undefined at x = 0. It is the full solution.

Where the marks go

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 if there are two.

The second mark needs to be visible on the paper. A sign diagram, or three substitutions written out, earns it; an answer that appears with no working does not, even when it is right.

Why multiplying through is not allowed

Polynomial inequalities, up to degree three

On the GDC: reading an inequality off a graph

TI-Nspire CX II

Casio fx-CG50

Your turn

Want a verdict on your own draft?