Introduction
Two graphs intersect when they pass through the same point. To prove that they intersect, it is enough to show that at some input their outputs are equal. The difficulty is that the point may not be easy to find by solving an equation directly.
Continuity provides a way to prove that an intersection exists without first calculating its exact location. The key step is to subtract the functions. Their difference measures which graph is higher, and a change in that difference's sign forces it to be zero somewhere in between.
Turn an Intersection into a Root
Suppose \(f\) and \(g\) are defined on an interval containing \([a,b]\). At a particular input \(x\), their graphs intersect if and only if \(f(x)=g(x)\). This equation is equivalent to \(f(x)-g(x)=0\).
The sign of \(h(x)\) records the graphs' vertical order at \(x\). If \(h(x)>0\), then \(f(x)>g(x)\); if \(h(x)<0\), then \(f(x)<g(x)\). Thus, if the sign switches from one endpoint to the other and \(h\) is continuous, the graphs must be equal somewhere between those endpoints.
The Intersection Theorem
Continuity of both functions on the same interval matters: it ensures that their difference \(h=f-g\) is continuous on that interval. The strict opposite-sign condition also matters. It says the two endpoint values of the difference lie on opposite sides of zero.
Why the Argument Works
The reasoning is a direct application of the Intermediate Value Theorem (IVT): a continuous function on a closed interval takes every value between its endpoint values. Here, zero lies strictly between \(h(a)\) and \(h(b)\), so the IVT guarantees an input where \(h\) takes the value zero.
Set \(h(x)=f(x)-g(x)\). Because \(f\) and \(g\) are continuous on \([a,b]\), their difference is continuous there.
Calculate \(h(a)\) and \(h(b)\). The hypothesis \(h(a)h(b)<0\) means one is negative and the other is positive.
Since zero lies between \(h(a)\) and \(h(b)\), some \(c\in(a,b)\) satisfies \(h(c)=0\).
From \(h(c)=f(c)-g(c)=0\), it follows that \(f(c)=g(c)\). The graphs therefore intersect at \((c,f(c))\).
This also proves the theorem: continuity gives the IVT hypothesis for \(h\), and the opposite endpoint signs put zero between its endpoint values. The IVT supplies \(c\); the definition of \(h\) then turns the root into an intersection.
Worked Examples
Worked Example: Two Lines with an Exact Intersection
Show that the graphs of \(f(x)=2x+1\) and \(g(x)=5-x\) intersect on \([0,2]\), and find the intersection.
Both functions are linear, so they are continuous on \([0,2]\). Define their difference:
Evaluate it at the endpoints:
Since \((-4)(2)=-8<0\), the endpoint values have opposite signs. The theorem guarantees some \(c\in(0,2)\) with \(f(c)=g(c)\). In this example, solve \(h(c)=0\):
Check the result by evaluating both functions at \(4/3\):
The values agree, so the intersection is \(\left(\frac{4}{3},\frac{11}{3}\right)\). The endpoint sign test proves existence; solving the difference gives the exact point here.
Worked Example: A Square-Root Graph and a Line
Show that \(f(x)=\sqrt{x+1}\) and \(g(x)=x\) intersect on \([0,3]\), and verify an intersection input.
The radicand \(x+1\) is positive throughout \([0,3]\), so the square-root expression is defined there and continuous. The function \(g(x)=x\) is continuous as well. Define
At the endpoints,
These values have opposite signs, since \(1(-1)=-1<0\). The theorem guarantees an intersection input in \((0,3)\). To find one, solve \(\sqrt{x+1}=x\). Squaring requires attention to the sign: the left side is nonnegative, so a solution must have \(x\geq 0\), as is true on this interval. Then
The root \((1-\sqrt{5})/2\) is negative, so it is not in \([0,3]\). The other root \(c=(1+\sqrt{5})/2\) is positive and less than \(3\), since \(\sqrt{5}<5\). Check it in the original equation, not just the squared equation. First, \(c^2=c+1\), so \(c+1=c^2\). Because \(c>0\), this gives \(\sqrt{c+1}=\sqrt{c^2}=c\). Thus \(f(c)=g(c)\), as required.
Worked Example: A Cosine Graph and a Constant
Show that \(f(x)=\cos x\) and \(g(x)=1/2\) intersect on \([0,\pi/2]\).
The cosine function and the constant function are continuous on this closed interval. Their difference is
At the left endpoint,
At the right endpoint,
The product of the endpoint values is \(-1/4<0\). Therefore the IVT applied to \(h\) guarantees some \(c\in(0,\pi/2)\) with \(h(c)=0\), which means \(\cos c=1/2\). Indeed, \(c=\pi/3\) is such a point: it lies in \((0,\pi/2)\) and \(\cos(\pi/3)=1/2\). Substitution checks the equality of the two function values.
What the Theorem Does—and Does Not—Say
The result guarantees at least one intersection on the interval. It does not, by itself, locate the intersection exactly, or establish that there is only one. In the first two examples, finding a point explicitly required additional algebra; in the third, a standard cosine value supplied one. In other problems, the IVT may establish existence even when an exact solution is unavailable.
The opposite-sign condition is sufficient, but it is not necessary for an intersection. For example, two graphs may meet at an endpoint, or they may meet inside the interval while the difference has the same sign at both endpoints. The theorem's endpoint test is a useful guarantee when it applies, not a complete test for every possible intersection.
If a difference is already zero at \(a\) or \(b\), the graphs intersect at that endpoint: \(h(a)=0\) means \(f(a)=g(a)\), and likewise at \(b\). In that case, no strict sign change is needed to establish that endpoint intersection.
A Common Mistake: Mixing the Endpoints
The endpoint test compares the two functions at the same input: \(h(a)=f(a)-g(a)\) and \(h(b)=f(b)-g(b)\). Comparing \(f(a)\) with \(g(b)\), or \(f(b)\) with \(g(a)\), does not calculate the endpoint values of one difference function. A sign change in those mismatched comparisons cannot be used in the IVT argument.
Worked Example: Correcting a Mismatched-Endpoint Argument
Consider \(f(x)=x\) and \(g(x)=x+\frac{1}{5}\) on \([0,1]\). An incorrect argument might compare values at opposite endpoints and claim a sign change:
But these are not \(h(0)\) and \(h(1)\) for \(h(x)=f(x)-g(x)\). Calculate the actual difference and its endpoint values:
There is no opposite-sign endpoint pair. In fact, \(f(x)\) is always \(1/5\) less than \(g(x)\), so they do not intersect on the interval. The correct calculation shows exactly why the mismatched comparison gave a false conclusion.
Why This Method Is Useful
Solving \(f(x)=g(x)\) directly can produce a complicated equation, or one with no convenient exact solution. The difference-function method separates the task into a manageable proof: check continuity, calculate two endpoint differences, and apply the IVT if their signs are opposite. It proves that the graphs meet without relying on a sketch, a numerical approximation, or a guessed intersection.
For a careful argument, identify the common interval first, verify continuity there, and compute each difference using the same endpoint for both function values. Then state what the IVT supplies and translate \(h(c)=0\) back into \(f(c)=g(c)\). That final translation is what connects the theorem about a function's values to the original question about two graphs.
Check Your Understanding
For each question, focus on the difference function and the hypotheses that make the IVT applicable.
- State the continuity and endpoint conditions that guarantee two functions intersect on \([a,b]\).
- If \(h(x)=f(x)-g(x)\), explain why \(h(c)=0\) means the graphs of \(f\) and \(g\) intersect.
- For \(f(x)=x^2\) and \(g(x)=3x-1\) on \([0,2]\), calculate \(h(0)\) and \(h(2)\). What does the theorem guarantee?
- Does having the same sign for \(h(a)\) and \(h(b)\) prove that the graphs do not intersect? Explain.
- Why is comparing \(f(a)-g(b)\) with \(f(b)-g(a)\) not a valid endpoint sign test for the difference function?
