When Equal Endpoint Values Force a Horizontal Tangent
Suppose a function begins and ends at the same height over an interval. If its graph is continuous between those endpoints and differentiable at every interior point, then somewhere along the way its tangent must be horizontal. Rolle’s Theorem makes this intuitive idea precise.
The theorem is a special case of the relationship between a function’s overall change and its derivative. Its conclusion is an existence statement: at least one interior point has derivative zero. The hypotheses matter, and the conclusion does not promise that the point is unique.
Statement of Rolle’s Theorem
Each condition has a specific role. Continuity is required on the entire closed interval, including its endpoints. Differentiability is required at every point strictly between the endpoints; the theorem does not require derivatives at \(a\) or \(b\). Finally, the endpoint values must be equal. When all three conditions hold, the conclusion concerns a point strictly inside the interval.
A value \(c\) satisfying \(f'(c)=0\) is often called a Rolle point for this function and interval. Since the derivative gives the tangent’s slope, \(f'(c)=0\) means the tangent line at \(x=c\) is horizontal.
Why the Theorem Is True
The proof uses two standard facts. The Extreme Value Theorem says that a function continuous on a closed, bounded interval attains an absolute maximum and an absolute minimum there. Fermat’s theorem says that if a function has a local maximum or minimum at an interior point and is differentiable there, its derivative at that point is zero.
Here is the reason for Fermat’s theorem. Suppose \(f\) has a local maximum at an interior point \(c\) and is differentiable there. For \(x\) sufficiently close to \(c\), \(f(x)\leq f(c)\). For \(x>c\), the difference quotient \(\frac{f(x)-f(c)}{x-c}\) is nonpositive. For \(x<c\), it is nonnegative. Since the derivative exists, the left- and right-hand limits of these quotients are equal. The common limit must therefore be both nonnegative and nonpositive, so it is zero. A local minimum follows by reversing the inequalities.
Continuity on \([a,b]\) means the Extreme Value Theorem guarantees that \(f\) attains both an absolute maximum and an absolute minimum on that interval.
If \(f\) is constant on \([a,b]\), its derivative is zero at every point of \((a,b)\), so the conclusion holds.
If \(f\) is not constant, there is some point where its value differs from the common endpoint value. If that value is larger, an absolute maximum occurs in the interior; if it is smaller, an absolute minimum occurs in the interior.
At the interior extremum, \(f\) is differentiable, so Fermat’s theorem gives \(f'(c)=0\) for an interior point \(c\).
This proves Rolle’s Theorem. In the nonconstant case, the key point is that the common endpoint value cannot be both the absolute maximum and the absolute minimum: the function takes some other value inside the interval. An extremum with that more extreme value cannot occur at either endpoint, so it must occur inside.
Finding a Rolle Point for a Quadratic
Worked Example: A Quadratic with Equal Endpoint Values
Apply Rolle’s Theorem to \(f(x)=x^2-4x+3\) on \([0,4]\), and find the point or points it guarantees.
First check continuity and differentiability. A polynomial is continuous and differentiable for every real \(x\), so the required conditions hold on \([0,4]\) and \((0,4)\).
Now evaluate the endpoints:
The endpoint values are equal, so Rolle’s Theorem guarantees at least one \(c\in(0,4)\) with \(f'(c)=0\). Differentiate and solve:
The value \(2\) is in \((0,4)\). Check the derivative and the endpoint values:
Thus the Rolle point is \(c=2\). In this example the derivative equation has just one solution in the interval, but the theorem itself guarantees at least one, not exactly one.
Finding a Rolle Point for a Trigonometric Function
Worked Example: Sine on a Half-Cycle
Apply Rolle’s Theorem to \(f(x)=\sin x\) on \([0,\pi]\).
The sine function is continuous on \([0,\pi]\) and differentiable on \((0,\pi)\). Its endpoint values are
Since the endpoints have equal values, Rolle’s Theorem guarantees an interior point \(c\) such that \(f'(c)=0\). Differentiate and solve:
The solutions of \(\cos c=0\) in \((0,\pi)\) give \(c=\frac{\pi}{2}\). This value is strictly between the endpoints. Check:
Therefore \(c=\frac{\pi}{2}\) is the Rolle point in this interval.
Solving for More Than One Rolle Point
Worked Example: A Cubic with Two Interior Points
Apply Rolle’s Theorem to \(f(x)=x^3-3x^2+2x\) on \([0,2]\), and find all points in \((0,2)\) where the derivative is zero.
As a polynomial, \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\). Evaluate both endpoints:
The hypotheses hold, so the theorem guarantees at least one zero of the derivative inside \((0,2)\). Differentiate and solve the derivative equation:
Using the quadratic formula,
Both solutions lie in \((0,2)\): since \(0<\frac{\sqrt{3}}{3}<1\), the smaller is between \(0\) and \(1\), and the larger is between \(1\) and \(2\). Check that each makes the derivative zero. Write \(c=1\pm\frac{1}{\sqrt{3}}\):
Thus both \(1-\frac{\sqrt{3}}{3}\) and \(1+\frac{\sqrt{3}}{3}\) are Rolle points. This example shows why it is important to solve the derivative equation fully: the theorem’s guarantee need not correspond to only one point.
What the Conclusion Does—and Does Not—Say
The equation \(f'(c)=0\) identifies an interior point where the graph has a horizontal tangent. It does not say that the function is zero at \(c\). For example, in the quadratic example, \(f(2)=-1\), even though \(f'(2)=0\). The theorem concerns the derivative, not the function’s value at the interior point.
Nor does the conclusion say that the function is constant or that the horizontal tangent occurs at the midpoint. Its location depends on the function. To find candidate points in a particular problem, differentiate and solve \(f'(x)=0\), then retain only solutions in the open interval \((a,b)\).
Rolle’s Theorem is useful because it turns information about endpoint values into a conclusion about an interior rate of change. Equal endpoint values mean the total change across the interval is zero; under the theorem’s smoothness conditions, the derivative must be zero somewhere inside.
A Common Pitfall: Equal Endpoints Are Not Enough
Worked Example: Why Differentiability Matters
Consider \(f(x)=|x|\) on \([-1,1]\). Its endpoint values are equal:
It would be incorrect to conclude from these equal values alone that Rolle’s Theorem applies. The function is continuous on \([-1,1]\), but it is not differentiable at \(x=0\), which lies inside the interval.
In fact, for \(x<0\), \(f'(x)=-1\), and for \(x>0\), \(f'(x)=1\). At zero, the left-hand difference quotient is \(-1\) and the right-hand difference quotient is \(1\), so \(f'(0)\) does not exist. There is no interior point where the derivative is zero.
The correction is to check every hypothesis before applying the theorem. Equal endpoint values and continuity do not replace the requirement of differentiability throughout the open interval.
Check Your Understanding
For each question, check the hypotheses before using the theorem, and give exact values where possible.
- State all three hypotheses of Rolle’s Theorem and its conclusion.
- For \(f(x)=x^2-6x+8\) on \([2,4]\), verify the endpoint condition and find the Rolle point.
- For \(f(x)=\cos x\) on \([-\pi,\pi]\), find the point or points in the open interval where the derivative is zero.
- Why does \(f(x)=|x|\) on \([-1,1]\) not satisfy all the hypotheses, even though its endpoint values agree?
- Does Rolle’s Theorem guarantee exactly one point with derivative zero? Explain what it guarantees instead.
