You already know how to solve $g(x) = f(x)$: it finds the exact inputs where two quantities are equal.
But most real questions are not about the tie, they are about who is winning: when does revenue beat cost, when does one dose stay above the safe threshold, when is one route faster than another.
That is exactly what an inequality like $g(x) \geq f(x)$ asks: for which values of $x$ is the curve $g$ on or above the curve $f$.
The good news is that you do not need a new formula for this, you need one clean habit and a way to track a sign.
Analogy
Picture two runners on the same track, their distance-from-start graphs drawn on one axis.
Solving $g(x) = f(x)$ finds the instants they are level, the moments one overtakes the other.
Solving $g(x) \geq f(x)$ finds the whole stretches of time when runner $g$ is ahead of or level with runner $f$.
Those stretches are intervals, and finding them is the entire skill of this subtopic.
One Move Powers It All: Compare The Difference To Zero
Comparing two changing functions directly is awkward, but comparing one function to the fixed number $0$ is easy.
So the first move, every single time, is to bring everything to one side: $$ g(x) \geq f(x) \quad\Longleftrightarrow\quad g(x) - f(x) \geq 0 $$
Give the difference a name, $h(x) = g(x) - f(x)$, and the question becomes a single question: where is $h(x)$ positive or zero?
The points of intersection of $g$ and $f$, which you meet in SL 2.4, are exactly the values where $h(x) = 0$: these are the boundary points.
Boundary points matter because a continuous function can only switch between positive and negative by passing through zero.
So between two neighbouring boundary points the sign of $h(x)$ never changes: it is all positive, or all negative.
That is the whole strategy: find the boundary points, then decide the sign of $h(x)$ on each interval between them.
Hint
Graphically, $h(x) \geq 0$ means the graph of $g$ sits on or above the graph of $f$, so the solution is simply the set of $x$ where $g$ is not below $f$.
Note
A continuous function changes sign only where it crosses zero, which is why the boundary points split the number line into fixed-sign pieces.
It can also flip sign where it is undefined, such as at the vertical asymptote of a rational function, so mark those break points too.
Test one point inside each piece and that single test decides the sign for the whole piece.
The Sign Diagram Method, Step By Step
The analytic method turns the strategy above into a reliable five-step routine.
Step 1: Rearrange
Move every term to one side so the inequality reads $h(x) \geq 0$ (or $\leq 0$), with $0$ alone on the right.
Step 2: Find the boundary points
Solve $h(x) = 0$ by factoring, and also note any $x$ where $h(x)$ is undefined.
Step 3: Draw the number line
Mark the boundary points in order; they cut the line into open intervals.
Step 4: Test each interval
Pick one easy value inside each interval, substitute it into $h(x)$, and record whether $h$ is positive or negative there.
Step 5: Read off and fix the endpoints
Keep the intervals whose sign matches, then include a boundary point if the inequality is $\geq$ or $\leq$, and exclude it if it is strict ($>$ or $<$) or the function is undefined there.
For the special case where $h$ is a quadratic, the same routine is the quadratic inequalities method from SL 2.7.
Example question
Solve $x^2 \geq x + 2$.
Solution
Rearrange to one side: $x^2 - x - 2 \geq 0$.
Factor the difference: $(x-2)(x+1) \geq 0$, so the boundary points are $x = -1$ and $x = 2$.
Test $x=-2$: $(-4)(-1) = 4 > 0$; test $x=0$: $(-2)(1) = -2 < 0$; test $x=3$: $(1)(4) = 4 > 0$.
The difference is positive outside the roots, and $\geq$ includes the roots, so the solution is $x \leq -1$ or $x \geq 2$.
Exam technique
There is no formula for inequalities in the formula booklet, so the method itself is what you must memorise, not recall from the booklet.
Marks are awarded for a clear sign analysis: show the factored difference and the tested signs, not just a final line.
Always state the boundary points explicitly before you write the interval, because an unsupported answer earns few method marks.
Write the final answer as an interval or an inequality in $x$, and double-check each endpoint against $\geq$ versus $>$.
A Cubic Inequality
A cubic can have up to three boundary points, so the number line can split into as many as four intervals.
The syllabus expects you to handle polynomials up to degree $3$ this way; the routine does not change, there are just more intervals to test.
Example question
Solve $x^3 \geq 4x$.
Solution
Rearrange: $x^3 - 4x \geq 0$.
Factor fully: $x(x^2 - 4) = x(x-2)(x+2) \geq 0$, so the boundary points are $x = -2,\; 0,\; 2$.
Test the four intervals: $x=-3$ gives $-15 < 0$; $x=-1$ gives $3 > 0$; $x=1$ gives $-3 < 0$; $x=3$ gives $15 > 0$.
Keep the positive intervals and include the zeros, giving $-2 \leq x \leq 0$ or $x \geq 2$.
Reading The Solution Straight Off A Graph
The graphical method skips the algebra and answers the question by eye, which is ideal when a sketch is given or a calculator is allowed.
Sketch or plot $y = g(x)$ and $y = f(x)$ on the same axes.
Mark every point where the two curves cross, because these are the same boundary points the algebra would find.
Scan along the $x$-axis and highlight the stretches where the graph of $g$ is on or above the graph of $f$.
Those highlighted stretches, written as intervals in $x$, are the solution.
Example question
The curves $y = g(x)$ and $y = f(x)$ meet only at $x = -1$ and $x = 3$. A sketch shows $g$ below $f$ for $x < -1$, above $f$ between $-1$ and $3$, and below $f$ again for $x > 3$. Solve $g(x) \geq f(x)$.
Solution
The boundary points are the crossings $x = -1$ and $x = 3$.
The graph of $g$ is on or above $f$ only on the middle stretch.
Since the inequality is $\geq$, include both endpoints: $-1 \leq x \leq 3$.
Hint
When you read a solution off a graph, the crossings are your only decision points, so pin them down first.
If two curves only touch (a repeated root) rather than cross, the sign of $g-f$ does not change there, which can make the solution a single point or all real numbers.
The Trap That Silently Flips The Inequality
The single biggest error in this topic is multiplying or dividing an inequality by an expression whose sign you do not know.
Multiplying $a \geq b$ by a positive number keeps the direction, but multiplying by a negative number reverses it to $\leq$.
So if you multiply by something like $(x-1)$, you cannot tell whether to flip the sign, because $(x-1)$ is positive for some $x$ and negative for others.
Cross-multiplying a fraction is the same mistake in disguise, and it is why you should never cross-multiply by a variable expression.
Dividing carries a second danger: dividing by an expression that could be zero can delete a valid boundary point, since the function is undefined there, not zero.
The cure is always the same: rearrange everything to one side against $0$, and let a sign diagram handle the changing signs for you.
Inequalities involving absolute values follow their own rules and are handled separately in AHL 2.16 on modulus inequalities.
Example question
Solve $\dfrac{3}{x-1} \geq 1$.
Solution
Do not multiply by $x-1$, since its sign is unknown; instead move everything to one side: $\dfrac{3}{x-1} - 1 \geq 0$.
Combine over a common denominator: $\dfrac{3 - (x-1)}{x-1} = \dfrac{4 - x}{x-1} \geq 0$.
Boundary points: the numerator is zero at $x = 4$ (included), and the expression is undefined at $x = 1$ (always excluded).
Test the intervals: $x=0$ gives $\tfrac{4}{-1} < 0$; $x=2$ gives $\tfrac{2}{1} > 0$; $x=5$ gives $\tfrac{-1}{4} < 0$.
Keep the positive interval, exclude $x=1$, include $x=4$: $1 < x \leq 4$.
Example question
Solve $x^2 > x + 2$, and compare your answer with the solution of $x^2 \geq x + 2$ found earlier.
Solution
Rearrange: $x^2 - x - 2 > 0$, so $(x-2)(x+1) > 0$ with boundary points $x = -1$ and $x = 2$.
The difference is positive outside the roots, exactly as before.
Because the inequality is now strict, the endpoints are excluded: $x < -1$ or $x > 2$.
The earlier $\geq$ version kept the endpoints, giving $x \leq -1$ or $x \geq 2$; only the endpoints differ.
Choosing Your Method: Sign Diagram, Graph, Or Technology
All three approaches answer the same question, so the skill is picking the one that fits the function in front of you.
Method
How it works
Best for
Main risk
Sign diagram
Factor g - f, mark zeros, test each interval
Polynomials you can factor (degree 2 or 3)
A wrong sign at a test point
Graph, above or below
Sketch g and f, read where g is on or above f
A given sketch or a quick check
Misreading a close or tangent crossing
Technology, GDC
Plot g - f, find the zeros, read the intervals
Functions you cannot factor by hand
Rounding the boundary points too early
Tip
If the difference factors cleanly, a sign diagram is fastest and gives exact boundary points.
If you are handed a sketch or only need a quick check, read the answer straight off the graph.
If the functions are not simple polynomials, use technology to locate the crossings, then read the intervals.
When Only Technology Will Do
When the boundary points are not nice numbers, factoring fails and the syllabus expects you to use technology to find them.
You still finish the same way: read off the intervals of $x$ where the difference is at or above zero, rounding only the final boundary points.
Example question
Solve $x^3 - 3x \geq 1$, giving your boundary points to $3$ significant figures.
Solution
Rearrange to one side: $x^3 - 3x - 1 \geq 0$.
This cubic does not factor over the rationals, so use a GDC to find the three zeros: $x \approx -1.53,\; -0.347,\; 1.88$.
A positive-leading cubic is on or above zero between the first two roots and beyond the largest root.
Reading the intervals: $-1.53 \leq x \leq -0.347$ or $x \geq 1.88$.
Active recall
What is the first move when solving $g(x) \geq f(x)$, and why does comparing to $0$ make the problem easier?
How do the points of intersection of $g$ and $f$ relate to the boundary points of the solution?
Why can the sign of $g(x) - f(x)$ only change at a boundary point or where the function is undefined?
Why must you never cross-multiply an inequality by an expression such as $(x-1)$?
When solving $g(x) \geq f(x)$, how do you decide whether a boundary point is included or excluded?
