Function Inequalities — AHL 2.15 Inequalities involving functions g(x) ≥ f(x) (graphical and analytical) (HL) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

Comparing functions is one of the oldest themes in mathematics. Long before formal algebra existed, ancient scholars needed to know when one quantity exceeded another — whether the supply of grain surpassed the demand, or whether one trajectory cleared a wall. The idea of function inequalities — asking when g(x) ≥ f(x) — grew naturally from these practical concerns and became a cornerstone of modern analysis, optimisation, and applied mathematics.

~300 BCE • Euclid's Geometric Comparisons

In Elements, Euclid compared magnitudes (lengths, areas) to establish when one exceeded another — an early, geometric form of inequality reasoning.

1637 • Descartes & Coordinate Geometry

René Descartes linked algebra to geometry, making it possible to plot curves on axes and visually compare two functions by examining which graph sits higher.

1821 • Cauchy Formalises Inequalities

Augustin-Louis Cauchy published rigorous treatments of inequalities in his Cours d'analyse, laying the groundwork for the analytic methods you will use in this topic.

1960s–Today • Modern Applications

Function inequalities underpin fields from economics (cost vs. revenue curves) to engineering (stress vs. material strength). GDC technology now lets students explore these relationships visually in seconds.

The central question this lesson addresses is deceptively simple: given two functions f and g, for which values of x is g(x) ≥ f(x)? Answering this question requires both graphical intuition and algebraic precision — two skills that the IB HL curriculum expects you to develop hand in hand.

SECTION 2

Core Principles & Definitions

Before diving into techniques, you need a clear understanding of the foundational ideas behind function inequalities. These principles connect what you already know about solving equations to the richer question of solving inequalities between entire functions.

Intersection Points as Boundaries

The inequality g(x) ≥ f(x) changes truth value at points where g(x) = f(x). Finding these intersection points is always the first step.

The Difference Function

Define h(x) = g(x) − f(x). Then g(x) ≥ f(x) is equivalent to h(x) ≥ 0. This transformation simplifies many problems.

Graphical Interpretation

g(x) ≥ f(x) means the graph of g lies on or above the graph of f. Shade the region where this is true to visualise the solution set.

Interval Notation for Solutions

Solutions are expressed as unions of intervals. Use square brackets [ ] when the endpoint is included (equality holds) and round brackets ( ) when it is excluded or extends to infinity.

Domain Awareness

The inequality only makes sense where both f and g are defined. Always consider the domains of the original functions before stating your answer.

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KEY TAKEAWAY

Think of two functions as two runners on a track. The inequality g(x) ≥ f(x) asks: "During which time intervals is Runner g at least as far ahead as Runner f?" The moments they are side by side — the intersection points — are where the lead changes. Between those moments, you only need to check one test point to determine who is in front for the entire interval.

SECTION 3

Visual Explanation — Graphical Method

The graphical method is the most intuitive approach and is always available on your GDC. By plotting both functions on the same axes, you can directly see where one curve sits above the other. The diagram below shows f(x) = x² and g(x) = 2x + 3. The shaded region highlights where g(x) ≥ f(x), that is, where the line is on or above the parabola.

The violet parabola represents f(x) = x² and the cyan line represents g(x) = 2x + 3. The yellow dots mark the intersection points at (−1, 1) and (3, 9). The shaded region between the curves shows where g(x) ≥ f(x), namely x ∈ [−1, 3].

From the diagram, you can read the solution directly: the line sits on or above the parabola for −1 ≤ x ≤ 3. The two intersection points — found by solving x² = 2x + 3 — act as the boundaries of the solution interval. Between the intersections, a single test point (say, x = 0) confirms that g(0) = 3 > 0 = f(0), verifying that the line is indeed above the parabola throughout the interval.

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GDC Tip

On your GDC, graph both functions and use the intersect feature to find the boundary x-values. Then visually confirm which curve is higher between those x-values. This graphical approach is accepted in IB examinations when supported by clear sketches in your working.

SECTION 4

Mathematical Framework — The Analytical Method

The analytical (algebraic) method provides exact answers without relying on a graph. The strategy has three stages: rearrange, find critical values, and test intervals. Every function inequality you encounter in AHL 2.15 can be attacked with this framework.

REARRANGEMENT

g(x) ≥ f(x) ⟺ g(x) − f(x) ≥ 0 ⟺ h(x) ≥ 0

Define h(x) = g(x) − f(x). The problem reduces to finding where the single function h is non-negative.

CRITICAL VALUES

Solve h(x) = 0 → x = x₁, x₂, …, xₙ

These roots of h split the real number line into intervals. The sign of h cannot change within an interval unless h passes through zero or is undefined.

SIGN ANALYSIS

Pick a test value c in each interval; if h(c) ≥ 0, the entire interval belongs to the solution.

By the Intermediate Value Theorem, a continuous function that is positive at one point and negative at another must cross zero between them. Since you have already found all zeros, the sign is constant on each sub-interval.

SOLUTION SET

S = { x ∈ Domain(f) ∩ Domain(g) : h(x) ≥ 0 }

Always express your final answer in interval notation, remembering to restrict to the intersection of the domains of f and g.

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Warning: Non-Polynomial Functions

When f or g involves logarithms, exponentials, or trigonometric functions, finding the roots of h(x) = 0 may not be possible algebraically. In those cases, use a combination of graphical and analytical methods: find the intersection points on the GDC, then confirm the sign of h in each interval analytically.

SECTION 5

Detailed Breakdown — Sign Diagrams & Interval Testing

A sign diagram (also called a sign chart or number-line analysis) is a visual tool that organises the test-point information. It places the critical values on a number line and records the sign of h(x) = g(x) − f(x) in each resulting interval. This makes it easy to read off the solution set at a glance.

The sign diagram for h(x) = −x² + 2x + 3. The yellow dots mark the critical values x = −1 and x = 3. A test point in each interval reveals whether h is positive (green, included) or negative (red, excluded).

The process shown above can be summarised in a table for quick reference. Each row in the table corresponds to one interval between consecutive critical values.

Interval	Test Point	h(x) Value	Sign	Include?
(−∞, −1)	x = −2	−(4) + (−4) + 3 = −5	Negative	No
[−1, 3]	x = 0	0 + 0 + 3 = 3	Positive	Yes ✓
(3, ∞)	x = 5	−25 + 10 + 3 = −12	Negative	No

Sign analysis for h(x) = −x² + 2x + 3

ℹ

Strict vs Non-Strict Inequalities

For g(x) ≥ f(x) (non-strict), include the endpoints where h(x) = 0 — use square brackets. For g(x) > f(x) (strict), exclude them — use round brackets. This detail costs easy marks on IB exams if overlooked.

SECTION 6

Worked Example — Full Analytical Solution

Let's work through a complete example involving non-polynomial functions. Find all values of x for which eˣ ≥ x + 1, i.e., determine where the exponential function is at least as large as the linear function g(x) = x + 1.

Solve eˣ ≥ x + 1 for x ∈ ℝ

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Step 1 — Form the Difference Function

Let h(x) = eˣ − (x + 1) = eˣ − x − 1. We need to find where h(x) ≥ 0. Both eˣ and x + 1 are defined for all real numbers, so the domain is ℝ.

h(x) = eˣ − x − 1, domain: x ∈ ℝ
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Step 2 — Find the Critical Values (Roots of h)

Set h(x) = 0: eˣ − x − 1 = 0, so eˣ = x + 1. We know that at x = 0, e⁰ = 1 and 0 + 1 = 1, so h(0) = 0. This means x = 0 is a root. To check whether there are other roots, we examine h'(x) = eˣ − 1. Setting h'(x) = 0 gives eˣ = 1, so x = 0 is the only critical point of h. Since h''(x) = eˣ > 0 for all x, the function h is concave up everywhere, so x = 0 is a minimum.

Only critical value: x = 0 (a minimum of h)
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Step 3 — Sign Analysis

Since h has a global minimum at x = 0 and h(0) = 0, the function h is non-negative everywhere. We can verify: h(−1) = e⁻¹ − (−1) − 1 = e⁻¹ ≈ 0.368 > 0, and h(1) = e¹ − 1 − 1 = e − 2 ≈ 0.718 > 0. The difference function h(x) touches zero at x = 0 and is strictly positive everywhere else.

h(x) ≥ 0 for all x ∈ ℝ
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Step 4 — State the Solution

Since h(x) = eˣ − x − 1 ≥ 0 for all real numbers, the inequality eˣ ≥ x + 1 holds for every x. The line y = x + 1 is the tangent to the curve y = eˣ at the point (0, 1), and the concavity of the exponential ensures the curve never dips below this tangent.

Solution: x ∈ ℝ (all real numbers), with equality only at x = 0

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INSIGHT

This example illustrates a special case: sometimes the inequality holds everywhere. The key was recognising that h has a global minimum of zero. In general, when the difference function is always non-negative (or always non-positive), the solution is the entire domain — or the empty set.

SECTION 7

Graphical vs Analytical — Strengths & Limitations

Both the graphical and analytical methods have their place in your IB toolkit. Understanding when to use each one — and when to use them together — will help you work more efficiently and score more marks on examinations.

Criterion	Graphical Method	Analytical Method
Precision	Approximate — limited by screen resolution and scale	Exact — gives solutions in closed form when algebraically solvable
Speed	Fast for an overview; use the GDC 'intersect' tool	Can be slow for complex functions, but gives rigorous proof
Applicability	Works for any pair of functions that can be graphed	Requires h(x) = 0 to be solvable — may need numerical methods otherwise
IB Exam Use	Accepted if sketch is shown with labelled points	Expected for 'show that' or 'prove' questions
Risk of Error	Misreading scale, missing intersections outside the window	Algebraic mistakes, forgetting domain restrictions

Comparison of graphical and analytical approaches to function inequalities

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BEST PRACTICE

The most reliable strategy is to use both methods together. Start with a graph to see the big picture — how many intersections there are, which function is on top, and whether the inequality holds on one interval or several. Then confirm with algebra for the exact boundary values. This combined approach protects you against both algebraic errors and misleading graphing windows.

SECTION 8

Connections to Advanced Theory

Function inequalities do not exist in isolation; they connect to several topics you will encounter later in the HL course and beyond. Recognising these links deepens your understanding and prepares you for university-level mathematics.

This Lesson	Advanced Extension	Where You'll See It
Finding where g(x) ≥ f(x)	Optimisation — finding max/min of g(x) − f(x)	AHL 5.9 (Calculus optimisation)
Area between curves	∫[a,b] (g(x) − f(x)) dx gives the enclosed area	AHL 5.12 (Applications of integration)
Sign of h(x) = g(x) − f(x)	Differential inequalities and stability analysis	University differential equations
Interval notation solutions	Set-builder notation, measure theory	University real analysis

How function inequalities connect to future topics

Perhaps the most powerful forward connection is to integration. When you know that g(x) ≥ f(x) on [a, b], the integral ∫ₐᵇ [g(x) − f(x)] dx gives the exact area of the region between the two curves. The sign analysis you learn in this topic is therefore a prerequisite for correctly setting up area integrals — you must always integrate the upper function minus the lower function.

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Looking Ahead

In the IA (Internal Assessment), function inequalities arise naturally in modelling. For instance, you might compare a model's predicted values g(x) with actual data f(x), and an inequality analysis tells you for which input ranges the model over- or under-estimates. This is a sophisticated application that examiners love to see.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain in your own words why solving g(x) ≥ f(x) is equivalent to solving h(x) ≥ 0, where h(x) = g(x) − f(x). Why is this rearrangement useful?

Show answer

Subtracting f(x) from both sides preserves the inequality (subtraction is valid for all real-valued functions). The resulting condition h(x) ≥ 0 reduces a comparison of two functions to a sign analysis of a single function. This is useful because we only need to find the zeros of one function — h — and determine its sign in each interval, rather than comparing two separate curves point by point. The technique also connects directly to the graphical interpretation: h(x) ≥ 0 means the graph of h lies on or above the x-axis.

Question 2: BASIC CALCULATION

Solve x² ≤ 4x − 3, expressing your answer in interval notation.

Show answer

Rearrange: x² − 4x + 3 ≤ 0. Factor: (x − 1)(x − 3) ≤ 0. The critical values are x = 1 and x = 3. Test x = 0: (−1)(−3) = 3 > 0 (outside). Test x = 2: (1)(−1) = −1 < 0 (inside). Test x = 4: (3)(1) = 3 > 0 (outside). Since we need the product ≤ 0, and the endpoints give equality, the solution is x ∈ [1, 3]. Graphically, this is the interval where the parabola y = x² lies on or below the line y = 4x − 3.

Question 3: INTERMEDIATE

Find all values of x for which ln(x) ≥ x − 1, where ln denotes the natural logarithm. State the domain and use calculus-based reasoning or a sign diagram.

Show answer

The domain of ln(x) is x > 0. Let h(x) = ln(x) − (x − 1) = ln(x) − x + 1. Then h'(x) = 1/x − 1. Setting h'(x) = 0 gives x = 1. Since h''(x) = −1/x² < 0 for all x > 0, x = 1 is a maximum. Evaluating h(1) = ln(1) − 1 + 1 = 0, the maximum value of h is zero. This means h(x) ≤ 0 for all x > 0, with equality only at x = 1. Therefore ln(x) ≥ x − 1 only when x = 1. The solution is x = 1 (a single point). This is actually the well-known inequality ln(x) ≤ x − 1.

Question 4: APPLIED

A company's revenue is modelled by R(x) = −2x² + 20x (in thousands of dollars) and its cost by C(x) = 4x + 6, where x is the number of units sold (in thousands). Find the range of x for which the company makes a non-negative profit, i.e., R(x) ≥ C(x).

Show answer

Profit is non-negative when R(x) ≥ C(x), i.e., −2x² + 20x ≥ 4x + 6. Rearranging: −2x² + 16x − 6 ≥ 0, or equivalently x² − 8x + 3 ≤ 0 (dividing by −2 and reversing the inequality). Using the quadratic formula: x = (8 ± √(64 − 12)) / 2 = (8 ± √52) / 2 = 4 ± √13. So x₁ = 4 − √13 ≈ 0.394 and x₂ = 4 + √13 ≈ 7.606. Since x must also be non-negative (units sold), the solution is x ∈ [4 − √13, 4 + √13], approximately [0.394, 7.606]. In context, the company profits when selling between about 394 and 7606 units.

Question 5: CRITICAL THINKING

Let f(x) = sin(x) and g(x) = x for x ≥ 0. Show that f(x) ≤ g(x) for all x ≥ 0, i.e., sin(x) ≤ x. Use the difference function approach and explain why the inequality is strict for x > 0.

Show answer

Let h(x) = x − sin(x). Then h(0) = 0 − 0 = 0, and h'(x) = 1 − cos(x). Since cos(x) ≤ 1 for all real x, we have h'(x) ≥ 0 for all x. This means h is non-decreasing. For x > 0, cos(x) < 1 at least on some sub-intervals (indeed, cos(x) = 1 only at isolated points x = 2kπ), so h is strictly increasing overall on (0, ∞). Since h(0) = 0 and h is strictly increasing for x > 0, h(x) > 0 for all x > 0. Therefore x − sin(x) > 0, which gives sin(x) < x for x > 0, with equality only at x = 0. This classical inequality is used extensively in calculus, particularly in the proof that lim(x→0) sin(x)/x = 1.

SUMMARY

Lesson Summary

Summary

To solve a function inequality of the form g(x) ≥ f(x), begin by forming the difference function h(x) = g(x) − f(x) and rewrite the problem as h(x) ≥ 0. Find the critical values where h(x) = 0 — these are the intersection points of the original graphs. Use a sign diagram with test points to determine the sign of h in each interval, then express the solution in interval notation, remembering to use square brackets for ≥ (non-strict) and round brackets for > (strict).

The graphical method — plotting both functions and reading where one sits above the other — provides intuition and a quick check. The analytical method delivers exact answers and is required for proof-style questions. In practice, combining both approaches is the most reliable strategy. Always check the domain of both functions before finalising your answer, and connect your work to broader ideas like area between curves in integration.