From Equal Endpoint Values to a Horizontal Tangent
A function can begin and end at the same height while changing in between. Rolle’s Theorem says that, under specific smoothness conditions, the graph must have at least one interior point where its tangent is horizontal. The result turns information about two endpoint values into a conclusion about a derivative somewhere inside the interval.
This is an existence guarantee: it tells us that such a point is present, but not necessarily where it is or whether there is only one. In applications, we use the theorem to know that a zero of the derivative exists, then solve \(f'(x)=0\) to locate possible points.
Statement of Rolle’s Theorem
Each condition matters to the conclusion. Continuity is required on the whole closed interval, differentiability is required at every interior point, and the endpoint values must agree. The point \(c\) is strictly inside the interval: Rolle’s Theorem does not promise that a derivative is zero at an endpoint.
The geometric picture is simple. The line joining the two endpoint points on the graph is horizontal because their heights are equal. Rolle’s Theorem guarantees that somewhere inside, the graph has a tangent with the same horizontal direction. In terms of the derivative, the tangent slope at that point is zero.
The equation on the left is the slope of the secant line through the endpoint values. Since \(f(a)=f(b)\), its numerator is zero, and since \(a<b\), its denominator is nonzero. The theorem’s additional content is the guarantee that this zero slope occurs as a derivative at some interior point.
How to Apply the Theorem
To use the theorem efficiently, first identify the interval and the function. Verify the endpoint equality and the required regularity, then state what the theorem guarantees. If the task asks for the point or points, differentiate and solve \(f'(x)=0\), retaining only solutions in \((a,b)\).
Write down \(f\), \(a\), and \(b\), with \(a<b\).
Confirm continuity on \([a,b]\), differentiability on \((a,b)\), and \(f(a)=f(b)\).
Conclude that \(f'(c)=0\) for at least one \(c\in(a,b)\).
Solve \(f'(x)=0\), keep interior solutions, and substitute them back into the derivative.
For polynomials, continuity and differentiability hold for every real input. That makes the regularity conditions straightforward in polynomial examples, but the endpoint equality and the requirement that the solution lie strictly inside the interval still need to be checked.
Worked Example: A Quadratic with Equal Endpoint Values
Apply Rolle’s Theorem to \(f(x)=x^2-4x+3\) on \([1,3]\), and find the guaranteed point.
Step 1: Check the hypotheses. The function is a polynomial, so it is continuous on \([1,3]\) and differentiable on \((1,3)\). Evaluate both endpoints:
Thus \(f(1)=f(3)\), so Rolle’s Theorem applies and guarantees some \(c\in(1,3)\) with \(f'(c)=0\).
Step 2: Solve for the point. Differentiate and set the derivative equal to zero:
Step 3: Check the location and derivative. The value \(2\) lies in \((1,3)\), and
Therefore the guaranteed point is \(c=2\). This verifies the conclusion directly.
What the Zero Derivative Means
At an interior point \(c\) where \(f'(c)=0\), the tangent line has slope zero. Its equation is \(y-f(c)=f'(c)(x-c)\), so at such a point it becomes \(y=f(c)\), a horizontal line.
This interpretation follows from the meaning of the derivative as the instantaneous slope. When the derivative exists, it is the limiting slope of nearby secant lines:
If this limit equals zero, the instantaneous slope at \(c\) is zero. The graph is locally flat to first order at that point, although it need not be constant near \(c\). Rolle’s Theorem guarantees the existence of at least one such interior point under its hypotheses; solving the derivative equation identifies where the guarantee is realized.
Worked Example: A Trigonometric Function
Apply Rolle’s Theorem to \(f(x)=\cos x\) on \([0,2\pi]\), and find the interior point where the derivative is zero.
Step 1: Check continuity and differentiability. The cosine function is continuous and differentiable for every real \(x\), so it has the required regularity on this interval.
Step 2: Check endpoint values.
The endpoint values agree, so Rolle’s Theorem guarantees at least one \(c\in(0,2\pi)\) with \(f'(c)=0\).
Step 3: Solve the derivative equation. Since \(f'(x)=-\sin x\),
In the open interval \((0,2\pi)\), the solution is \(x=\pi\). The endpoint solutions \(0\) and \(2\pi\) are not interior points and are not included. Check the result:
Thus \(c=\pi\) is an interior point with a horizontal tangent.
More Than One Point Can Be Guaranteed
The phrase “there exists at least one” does not mean “there is exactly one.” A function may have several interior points where its derivative is zero. When asked to find the points in a particular example, solve the derivative equation completely on the specified open interval; do not stop after finding one valid point.
Worked Example: Two Interior Points
Apply Rolle’s Theorem to \(f(x)=x^3-3x\) on \([-\sqrt{3},\sqrt{3}]\), then find all interior solutions of \(f'(x)=0\).
Step 1: Check the hypotheses. This is a polynomial, so it is continuous on the closed interval and differentiable on the open interval. At either endpoint, \(x^2=3\), so \(x^3-3x=x(x^2-3)=0\). In particular,
The endpoint values are equal. Rolle’s Theorem guarantees at least one point \(c\in(-\sqrt{3},\sqrt{3})\) with \(f'(c)=0\).
Step 2: Find the derivative’s zeros.
Set the derivative equal to zero and solve:
Step 3: Check both points. Since \(\sqrt{3}>1\), both values lie strictly between \(-\sqrt{3}\) and \(\sqrt{3}\). Substitution confirms:
There are two interior points with zero derivative, \(c=-1\) and \(c=1\). Both meet the theorem’s conclusion.
A Common Mistake: Using Endpoint Equality Alone
Equal endpoint values are not enough by themselves. The differentiability condition cannot be omitted. For example, consider \(g(x)=|x|\) on \([-1,1]\). The endpoint values agree:
A mistaken argument might say that Rolle’s Theorem therefore guarantees a \(c\in(-1,1)\) with \(g'(c)=0\). But \(g\) is not differentiable at \(x=0\). At every other interior point its derivative is either \(-1\) or \(1\):
There is no interior point where \(g'(x)=0\). The correction is to check all of the theorem’s hypotheses before using its conclusion. Here the missing differentiability condition means the theorem does not apply; the example does not contradict Rolle’s Theorem.
Using the Guarantee Carefully
Rolle’s Theorem is useful whenever endpoint information is available but the interior behavior is not yet known. It can assure us that a derivative equation has an interior solution even before we calculate the solution explicitly. In worked problems, differentiating then provides a way to locate such points and check them.
When interpreting a result, keep the theorem’s scope in view. It guarantees a horizontal tangent, not that the graph is horizontal throughout the interval. If the function is constant, every interior point has zero derivative; otherwise, there may be one or several such points. In either case, the conclusion concerns at least one point strictly between the endpoints.
Check Your Understanding
Use the theorem’s conditions and conclusion to answer these questions.
- State the three hypotheses and the conclusion of Rolle’s Theorem.
- For \(f(x)=x^2-6x+8\) on \([2,4]\), what must you verify before applying the theorem?
- Why must a point \(c\) supplied by Rolle’s Theorem lie in the open interval rather than at an endpoint?
- For \(f(x)=x^3-3x\) on \([-\sqrt{3},\sqrt{3}]\), which interior points have zero derivative?
- Why does the equality \(g(-1)=g(1)\) not allow Rolle’s Theorem to be applied to \(g(x)=|x|\) on \([-1,1]\)?
