Why the Order of the Curves Matters
When describing a region between two graphs, it is not enough to know which functions form its boundaries. On a vertical slice, the upper curve is the one with the greater \(y\)-value, and the lower curve is the one with the smaller \(y\)-value. Identifying them correctly gives the slice the right height.
The comparison is made at the same \(x\)-value. A graph may make the order easy to see when the curves are far apart, but visual inspection becomes unreliable when the curves are close, nearly parallel, or intersect. Function values provide an exact check.
These terms are relative to the specified interval. A function that is above another on one interval may be below it on a different interval. If the order changes inside an interval, neither function is upper throughout that entire interval.
Comparing Function Values
To compare two curves at a test point \(c\), evaluate both functions at \(c\). The larger value gives the higher point on the vertical line \(x=c\). It is also useful to subtract one function from the other. For example, if \(f(c)-g(c)>0\), then \(f\) is above \(g\) at \(c\).
A test point establishes the order at that point. To use it to identify the order throughout an interval, we also need to know that the curves cannot switch order somewhere else in the interval. For continuous functions, the key fact is that a change in order requires an intersection.
Why One Test Point Can Be Enough
Here is the reason behind the theorem. Define the difference \(h(x)=f(x)-g(x)\). Because \(f\) and \(g\) are continuous, \(h\) is continuous. If \(f\) is above \(g\) at a point \(a\), then \(h(a)>0\). If \(f\) were below \(g\) at another point \(b\), then \(h(b)<0\). The Intermediate Value Theorem would then imply that \(h(c)=0\) for some point \(c\) between \(a\) and \(b\). At that point, \(f(c)=g(c)\), so the curves intersect. Therefore, if there is no intersection, the difference cannot change sign, and the order found at one test point holds throughout the interval.
This argument requires continuity. It also requires that the interval contain no point where the functions are equal if we want a strict ordering everywhere. If the graphs meet at an endpoint, the curves may still have a consistent order in the interior, with equality at that endpoint. If they meet inside the interval, test points on opposite sides may give different answers.
Worked Example: Comparing Close Curves on an Interval
Let \(f(x)=1.01x+2\) and \(g(x)=x+2\) on \([0,3]\). Determine which curve is upper on this interval.
First compare their values at the test point \(x=1\):
Thus \(f(1)>g(1)\). To check whether this order persists, subtract:
For \(0<x\le 3\), \(0.01x>0\), so \(f(x)>g(x)\). At the left endpoint, \(f(0)=2=g(0)\). Therefore \(f\) is the upper curve and \(g\) the lower curve throughout the interval, with equality at \(x=0\). The difference is small: at \(x=3\), it is \(0.01(3)=0.03\). This is why the graph alone may not make the ordering obvious.
Check the endpoint comparisons: \(f(0)=g(0)=2\), and \(f(3)=5.03\) while \(g(3)=5\), consistent with the conclusion.
Intervals With a Known Intersection
An intersection occurs where \(f(x)=g(x)\). If an intersection is inside the specified interval, the order can change there. A single test point for the whole interval may then be misleading: it tells you the order at that point, not necessarily everywhere.
When an intersection is already known, compare the functions at test points on the relevant sides of it. The following example uses a specified interval that lies entirely on one side of an intersection.
Worked Example: An Interval Beside an Intersection
Let \(f(x)=x\) and \(g(x)=x^2\) on \([0,\tfrac12]\). The functions meet at \(x=0\), since \(f(0)=0=g(0)\). Determine their order on the interval.
Choose the interior test point \(x=\tfrac14\). Then
Since \(\frac14=\frac4{16}>\frac1{16}\), \(f\) is above \(g\) at the test point. To verify the order over the interval, use the difference:
For \(0<x\le \tfrac12\), both \(x>0\) and \(1-x\ge \tfrac12>0\). Hence \(x(1-x)>0\), so \(f(x)>g(x)\) in the interior. At \(x=0\) the values are equal. Thus \(f\) is upper and \(g\) lower on the interval, with a meeting point at the left endpoint.
Check the other endpoint: \(f(\tfrac12)=\tfrac12=\tfrac8{16}\), whereas \(g(\tfrac12)=\tfrac14=\tfrac4{16}\). The comparison agrees with the conclusion.
When the Order Changes Inside the Interval
If the curves intersect inside the interval, first establish that intersection locations are relevant to the comparison. Then test points on the different sides. The order may be consistent on each side but different across the intersection. In that situation, it is inaccurate to name one curve as the upper curve on the entire interval.
Worked Example: A Crossing Within the Specified Interval
Compare \(f(x)=x^2\) and \(g(x)=1\) on \([-2,2]\). Their graphs intersect when \(x^2=1\), which gives \(x=-1\) and \(x=1\).
Test the point \(x=0\), which lies between the intersections:
Here \(g(0)>f(0)\). Now test \(x=-\tfrac32\), which lies to the left of \(-1\):
Thus \(f\) is above \(g\) at that point. A test on the right gives the same ordering: \(f(\tfrac32)=\tfrac94>1=g(\tfrac32)\). Consequently, the order is not uniform across \([-2,2]\): \(g\) is above \(f\) between \(-1\) and \(1\), while \(f\) is above \(g\) to the left of \(-1\) and to the right of \(1\). They are equal at the intersection points.
Check the transition values directly: \(f(-1)=1=g(-1)\) and \(f(1)=1=g(1)\). Each change of order occurs at a point where the curves meet. The conclusion concerns which graph is higher; it does not assign one upper curve to the full interval.
A Reliable Comparison Procedure
Use a graph to see the likely arrangement, but base the final identification on function values and the interval. If the curves are continuous and known not to intersect inside the interval, one well-chosen test point is enough to establish their order throughout. If intersections are present, test on the sides of those points rather than assuming the order is uniform.
The same pair of functions may have different relative order on different intervals.
Use the graph to form a hypothesis, but do not rely on visual closeness or apparent separation as proof.
Evaluate both functions at the same \(x\), or compute \(f(x)-g(x)\).
For continuous functions with no intersection in the interval, it cannot. If there is an interior intersection, compare on its different sides.
Say which curve is upper on which interval, and include any points where the curves are equal.
Common Pitfall: Trusting One Test Point Without Checking Intersections
A test point does not automatically prove the order over an entire interval. For example, in the preceding example, at \(x=0\) we found \(g(0)>f(0)\). The incorrect conclusion would be that \(g\) is the upper curve throughout \([-2,2]\). But at \(x=\tfrac32\),
So the claim fails. The correction is to notice the known intersections at \(-1\) and \(1\), then test on the different sides: \(g\) is higher between the intersections, and \(f\) is higher outside them within the specified interval. A test point gives a local comparison; continuity and the intersection information determine where that comparison remains valid.
Why This Identification Is Useful
In a vertical slice of a region between graphs, the slice height is found by taking the \(y\)-coordinate at the top and subtracting the \(y\)-coordinate at the bottom. Correctly identifying the upper and lower curves is therefore essential when setting up an area calculation. Reversing them reverses the sign of the difference.
The central habit is simple: compare values at the same input, and then ask whether the order can change elsewhere in the interval. Graphs suggest; exact comparisons and intersection information justify.
Check Your Understanding
For each question, support your conclusion with a comparison of function values or a difference of functions.
- Suppose \(f\) and \(g\) are continuous on \([a,b]\), have no intersection there, and \(f(c)>g(c)\) for one \(c\in[a,b]\). What can you conclude about their order throughout \([a,b]\), and why?
- For \(f(x)=3-x\) and \(g(x)=1\) on \([0,1]\), which curve is upper? Check at \(x=\tfrac12\) and explain why the order holds throughout.
- The graphs of two continuous functions meet at an interior point of an interval. Why is a test point on only one side insufficient to determine the order on the entire interval?
- For \(f(x)=x^2\) and \(g(x)=4\) on \([-3,3]\), compare their values at \(x=0\) and \(x=3\). What does this tell you about whether one curve is upper throughout the interval?
