This whole subtopic is higher level. Nothing in it is on an SL paper.
Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 2 Functions · 2.15 Solutions of g(x) ≥ f(x), graphically and analytically
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Rewrite g(x) ≥ f(x) as g(x) − f(x) ≥ 0 and solve it | HL | Every inequality question starts here |
| Solve polynomial inequalities up to degree 3 by factorising and a sign diagram | HL | "Solve x³ − 7x − 6 > 0" (4 marks), often after "factorise fully" |
| Solve the same inequalities from a graph, reading where one curve is above the other | HL | "Use the graph to solve…"; a sketch that shows the answer |
| Handle repeated roots, strict and non-strict inequalities, and excluded points | HL | The difference between x > −3 and x > −3, x ≠ 1 is the last mark |
| Solve rational inequalities without multiplying by an expression of unknown sign | HL | "Solve (2x − 1)/(x + 1) < 1" (4 marks) |
| Use technology for these and other functions | HL | Paper 2: "Solve eˣ ≥ x + 2" (3 marks), answers to 3 s.f. |
| Write the solution set correctly, with "or" between separate pieces | HL | Accuracy mark for the final set |
Before you start
You need the quadratic inequalities of 2.7 (solve the equation, sketch the parabola, read off the region), the factor theorem and division from 2.12, and the rational function sketches of 2.13. From 2.4, the points where two graphs meet are the solutions of f(x) = g(x); that is the whole bridge between equations and inequalities.
1The idea in one paragraph
An equation f(x) = g(x) asks where two graphs meet. An inequality g(x) ≥ f(x) asks where the graph of g is on or above the graph of f. The meeting points split the x-axis into pieces, and on each piece one graph stays on top, because to swap over they would have to meet again. So the method is always the same: bring everything to one side, h(x) = g(x) − f(x) ≥ 0, find where h is zero (and where it is undefined), and then decide the sign of h on each piece between those critical values. With factors you decide the signs by algebra; without them you read them off a graph, drawn by hand or by the GDC.
2Two graphs, one question
Figure 1 shows f(x) = x² − 2x − 3 and g(x) = x + 1. Where is g(x) ≥ f(x)?
Graphically. The graphs meet at (−1, 0) and (4, 5). Between those two x-values the line is above the parabola; outside them it is below. The answer is the set of x-values, the amber bar on the x-axis: −1 ≤ x ≤ 4. The end points are included because the inequality is ≥ and the graphs are equal there.
Analytically.
The two methods are the same thought. Subtracting turns "g above f" into "g − f above the x-axis", and the x-axis is much easier to compare with than a second curve.
To solve g(x) ≥ f(x), solve h(x) = g(x) − f(x) ≥ 0: find the zeros of h, then the sign of h between them.
3Cubic inequalities and the sign diagram
The guide expects polynomials up to degree 3 by algebra. Factorise first, using the factor theorem if you need to; then build a sign diagram.
Worked example 1 (Paper 1). Solve x³ − 2x² − 5x + 6 ≤ 0.
Find a root by trial: P(1) = 1 − 2 − 5 + 6 = 0, so (x − 1) is a factor. Dividing out, x³ − 2x² − 5x + 6 = (x − 1)(x² − x − 6) = (x − 1)(x + 2)(x − 3).
The critical values are −2, 1 and 3. They cut the line into four intervals. In each one, every factor keeps its sign, so the product keeps its sign:
| Interval | x + 2 | x − 1 | x − 3 | Product |
|---|---|---|---|---|
| x < −2 | − | − | − | − |
| −2 < x < 1 | + | − | − | + |
| 1 < x < 3 | + | + | − | − |
| x > 3 | + | + | + | + |
We want ≤ 0: the negative intervals, plus the zeros themselves. The answer is x ≤ −2 or 1 ≤ x ≤ 3. Figure 2 shows the same thing as a graph, with the sign row beneath it.
A quicker way to fill the table: with three distinct single roots and a positive leading coefficient, the signs alternate from + on the far right (the cubic's right-hand end goes up), so reading leftwards they are +, −, +, −. That shortcut is safe only when every root is single, which is the next section.
4Repeated roots and excluded points
A squared factor never changes sign. So at a double root the graph touches the axis and turns back, and the sign on the two sides is the same.
Worked example 2. Solve (x − 1)²(x + 3) > 0.
(x − 1)² is positive everywhere except at x = 1, where it is 0. So the sign of the product is the sign of (x + 3), except at x = 1. We need it strictly greater than zero, so x = 1 must be removed:
x > −3, x ≠ 1
Figure 3 shows why. The graph crosses the axis at −3 and touches it at 1, where the value is exactly 0, which does not satisfy "> 0". The hollow circle marks the excluded point. Had the question said ≥ 0, the answer would be x ≥ −3, with 1 included.
Strict (<, >) and non-strict (≤, ≥) inequalities differ only at the critical values. Decide each end point separately: a zero of the numerator is included for ≤ or ≥; an asymptote (a zero of the denominator) is never included, because the function has no value there.
5Never divide by an expression that might be negative
The most common way to lose half the answer is an innocent-looking division.
Worked example 3 · the trap. Solve x³ ≥ 4x.
Dividing both sides by x gives x² ≥ 4, so x ≤ −2 or x ≥ 2. That is wrong. Dividing by x is only allowed when x > 0; when x < 0 it reverses the inequality, and when x = 0 it is not allowed at all. The correct method keeps everything as factors:
Figure 4 shows the piece the division lost: between −2 and 0, x³ is above 4x as well.
The same care goes for multiplying. Multiplying both sides by a negative number reverses the inequality, and since you usually do not know the sign of an expression like (x − 2), you must not multiply by it either. Squaring both sides is also unsafe unless both sides are known to be non-negative.
Worked example 4 · two polynomials. Solve x³ + x² ≥ 4x + 4.
6Rational inequalities
The guide lets you use technology for functions beyond polynomials, but a simple rational inequality is quick on Paper 1 as long as you keep the rule above: do not multiply by the denominator. Bring everything to one side as a single fraction, and then use a sign diagram on the top and bottom together.
Worked example 5 (Paper 1). Solve (x + 1)/(x − 2) ≥ 3.
| Interval | 7 − 2x | x − 2 | Fraction |
|---|---|---|---|
| x < 2 | + | − | − |
| 2 < x < 3.5 | + | + | + |
| x > 3.5 | − | + | − |
The fraction is ≥ 0 on 2 < x < 3.5, and equal to 0 at x = 3.5. At x = 2 it is undefined, so 2 is excluded. The answer is 2 < x ≤ 3.5. Figure 5 confirms it: the curve sits on or above the line y = 3 only between the asymptote and the meeting point.
Had you multiplied both sides by (x − 2), you would have got x + 1 ≥ 3x − 6, so x ≤ 3.5: every x less than 2 included, all of them wrong. An alternative that is safe is to multiply by (x − 2)², which is positive, and then factorise; it gives the same answer with more algebra.
7With technology
For anything that will not factorise, including exponentials, logarithms, trigonometric functions or a cubic with awkward roots, Paper 2 expects the GDC. The method is exactly the graphical method of section 2.
Worked example 6 (Paper 2). Solve eˣ ≥ x + 2.
- Graph y = eˣ and y = x + 2 in a window wide enough to show every meeting point. Or graph y = eˣ − x − 2 and look for its zeros.
- Use the intersect tool at each meeting point: x = −1.84 and x = 1.15 (3 s.f.).
- Read the order of the graphs on each piece. Figure 6 shows eˣ on top to the left of −1.84 and to the right of 1.15.
- Write the set: x ≤ −1.84 or x ≥ 1.15.
Two checks before you write the answer. Zoom out to make sure there are no more meeting points off the screen. And look for points where either function is undefined, such as a logarithm's domain or an asymptote, since the answer can also change there.
8Writing the answer
- Use "or" between separate pieces: x ≤ −2 or 1 ≤ x ≤ 3. Writing "x ≤ −2 and 1 ≤ x ≤ 3" describes no numbers at all.
- A compound inequality such as −1 ≤ x ≤ 4 is one piece. Never write 4 ≤ x ≤ −1 or −2 ≥ x ≥ 3.
- Interval notation is equally acceptable: [−1, 4], or ]−∞, −2] ∪ [1, 3], with square brackets turned outwards (or round brackets) for excluded ends, as the IB writes them.
- State excluded single points: x > −3, x ≠ 1.
- On Paper 2, give critical values to 3 s.f. and keep the inequality signs from the question.
9Where marks are lost
Dividing or multiplying by an expression of unknown sign. x³ ≥ 4x divided by x loses −2 ≤ x ≤ 0. (x + 1)/(x − 2) ≥ 3 multiplied by (x − 2) gains every x below 2. Factorise instead.
Solving the equation and stopping. Finding the critical values is half the work. The answer is a set of intervals, not a list of roots.
Alternating signs through a repeated root. A squared factor does not change sign. At a double root the sign is the same on both sides.
Including an asymptote. Where the denominator is zero, the function has no value, so the point is never in the solution set, even for ≥.
Forgetting to exclude a double root for a strict inequality. (x − 1)²(x + 3) > 0 fails at x = 1, where the value is 0.
Writing "and" instead of "or". Two separate pieces of the line are joined by "or", or by ∪.
Missing a meeting point on the GDC. Zoom out. An exponential and a line can meet twice, and the second meeting is often off the default screen.
Rounding critical values early. Give 3 s.f. in the answer, but use the stored value if the result feeds another part.
10Work it right
- Bring every term to one side: h(x) = g(x) − f(x), compared with 0.
- Factorise h fully (factor theorem for a cubic), or write a rational h as one fraction.
- List the critical values: zeros of the top, and zeros of the bottom.
- Build a sign diagram, or sketch h; alternate signs only through single roots.
- Pick the intervals the inequality asks for; include zeros for ≤ or ≥, never include asymptotes.
- On Paper 2, graph both sides, intersect at every meeting point, and read which is on top.
- Write the set with "or" between pieces and any excluded points stated.
11Try it
Marks in brackets. Q1 to Q4 are Paper 1 style, no calculator. Q5 is Paper 2 style, with a GDC.
Q1. Solve the inequality x³ − 7x − 6 > 0. 4 marks
Q2. Solve the inequality 2x³ ≤ x² + x. 5 marks
Q3. Solve the inequality (x − 3)²(2 − x) < 0. 3 marks
Q4. Solve the inequality (2x − 1)/(x + 1) < 1. 4 marks
Q5. Solve the inequality ln(x + 3) > x² − 1. 4 marks
12In one breath
To solve g(x) ≥ f(x), look for where the graph of g is on or above the graph of f, which is the same as where h = g − f is on or above the x-axis. The meeting points, and any points where a function is undefined, are the critical values; between them the sign of h cannot change. For polynomials up to degree 3, factorise (the factor theorem finds a cubic's first factor), then use a sign diagram: signs alternate through single roots and stay the same through a squared factor. Never divide or multiply by an expression whose sign you do not know: x³ ≥ 4x divided by x loses a whole interval. For a rational inequality, make one fraction and sign the top and bottom together; an asymptote is never included. For anything else, use the GDC: intersect at every meeting point, zoom out, and read which graph is on top. Include end points for ≤ and ≥, exclude them for < and >, and join separate pieces with "or".
Answers
Q1. P(−1) = −1 + 7 − 6 = 0, so (x + 1) is a factor, and x³ − 7x − 6 = (x + 1)(x² − x − 6) = (x + 1)(x + 2)(x − 3). Critical values −2, −1, 3; with single roots and a positive leading coefficient the signs from the right are +, −, +, −. So the product is positive for −2 < x < −1 or x > 3. M1 for finding a factor, A1 for full factorisation, M1 for a sign diagram or sketch, A1 for the correct set with strict inequalities.
Q2. 2x³ − x² − x ≤ 0, so x(2x² − x − 1) ≤ 0, so x(2x + 1)(x − 1) ≤ 0. Critical values −½, 0, 1; signs from the right +, −, +, −. The answer is x ≤ −½ or 0 ≤ x ≤ 1. M1 for bringing everything to one side without dividing by x, A1 for the factorisation, M1 for the sign analysis, A1 for x ≤ −½, A1 for 0 ≤ x ≤ 1. Dividing by x at the start and giving only −½ ≤ x ≤ 1 or similar scores at most M0 A0 M1 A0 A0.
Q3. (x − 3)² ≥ 0, with equality only at x = 3, so the sign is that of (2 − x), except at x = 3 where the product is 0. We need it negative: 2 − x < 0, so x > 2, and x = 3 must be excluded. x > 2, x ≠ 3. M1 for recognising the squared factor does not change sign, A1 for x > 2, A1 for excluding x = 3.
Q4. (2x − 1)/(x + 1) − 1 < 0 gives (2x − 1 − x − 1)/(x + 1) < 0, that is (x − 2)/(x + 1) < 0. Critical values −1 and 2. The fraction is negative when the top and bottom have opposite signs, which is between them: −1 < x < 2. M1 for a single fraction, A1 for (x − 2)/(x + 1), M1 for the sign analysis, A1 for the set. Multiplying through by (x + 1) and giving x < 2 scores M0 A0 M0 A0.
Q5. The domain needs x > −3. Using the GDC, y = ln(x + 3) and y = x² − 1 meet at x = −1.25 and x = 1.59 (3 s.f.), and the logarithm is above the parabola between them. −1.25 < x < 1.59. M1 for a graphical approach with both functions (or their difference), A1 for −1.25, A1 for 1.59, A1 for the set with strict inequalities. Near x = −3 the logarithm falls to −∞ while the parabola is near 8, so there is no further piece.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 2.15 Solutions of g(x) ≥ f(x), graphically and analytically. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.