Determine Whether Rolle's Theorem Can Be Applied

Rolle's theorem is a fundamental result incalculus that provides a simple guarantee about the existence of a stationary point for a function that meets three specific conditions. Understanding how to verify those conditions is essential for students who want to apply the theorem correctly in homework problems, exams, or real‑world modeling situations. This article walks you through the theorem’s statement, breaks down each requirement, offers a practical checklist, illustrates the process with several examples, highlights common pitfalls, and answers frequently asked questions.

Understanding Rolle's Theorem

Rolle's theorem states: If a function (f) is continuous on the closed interval ([a,b]), differentiable on the open interval ((a,b)), and satisfies (f(a)=f(b)), then there exists at least one number (c) in ((a,b)) such that (f'(c)=0).

In plain language, when a smooth curve starts and ends at the same height over an interval, somewhere inside that interval the curve must have a horizontal tangent (a point where the derivative equals zero). The theorem is a special case of the Mean Value Theorem and serves as a building block for many proofs in analysis.

Conditions for Applying Rolle's Theorem

To decide whether Rolle's theorem can be applied to a given function on a specified interval, you must check three distinct conditions:

1. Continuity on ([a,b]) – The function must have no breaks, jumps, or vertical asymptotes anywhere from (a) to (b), inclusive.
2. Differentiability on ((a,b)) – The function must possess a derivative at every interior point; corners, cusps, or vertical tangents disqualify differentiability at those points.
3. Equal endpoint values – The function’s values at the two endpoints must be identical: (f(a)=f(b)).

If any one of these fails, Rolle's theorem does not guarantee the existence of a point with zero derivative, although such a point might still exist by coincidence.

Quick Reference Checklist

• [ ] Is (f) continuous on the whole closed interval ([a,b])?
• [ ] Is (f) differentiable on the whole open interval ((a,b))?
• [ ] Do we have (f(a)=f(b))? Only when all three boxes are checked can you confidently invoke Rolle's theorem.

Step‑by‑Step Guide to Verify the Conditions Follow these steps whenever you encounter a problem that asks you to “determine whether Rolle's theorem can be applied.”

Step 1: Identify the interval ([a,b]) and the function (f(x))

Write down the exact interval and the explicit formula (or piecewise definition) of the function.

Step 2: Test continuity on ([a,b])

• Look for points where the denominator of a rational expression equals zero.
• Check for piecewise definitions: ensure the pieces join without a jump at the boundaries.
• Verify that any radicals, logarithms, or trigonometric components are defined throughout the interval.
• If you find a discontinuity, note its location; the theorem cannot be applied.

Step 3: Test differentiability on ((a,b))

• Differentiate the function symbolically (if possible) and see where the derivative might be undefined.
• Common trouble spots: corners (absolute value), cusps (e.g., (x^{2/3}) at (x=0)), vertical tangents (e.g., (\sqrt[3]{x}) at (x=0)), and points where the original function is not continuous.
• If the derivative fails to exist at any interior point, differentiability fails.

Step 4: Compare endpoint values

• Compute (f(a)) and (f(b)).
• If they are exactly equal (or differ only by a negligible rounding error in an applied context), the condition holds.

Step 5: Conclude

• If all three steps succeeded, state that Rolle's theorem applies and therefore there exists at least one (c\in(a,b)) with (f'(c)=0).
• If any step failed, explain which condition is violated and conclude that the theorem cannot be applied (though you may still search for a stationary point by other means).

Worked Examples

Example 1: Polynomial Function

Problem: Determine whether Rolle's theorem can be applied to (f(x)=x^{3}-3x+2) on ([-2,2]).

Solution:

1. Continuity: Polynomials are continuous everywhere → condition satisfied.
2. Differentiability: Polynomials are differentiable everywhere → condition satisfied.
3. Endpoint values:
   (f(-2)=(-2)^{3}-3(-x)+2=-8+6+2=0)
   (f(2)=2^{3}-3(2)+2=8-6+2=4)
   Since (f(-2)\neq f(2)), the third condition fails.

Conclusion: Rolle's theorem cannot be applied because the function values at the endpoints differ.

Example 2: Trigonometric Function

Problem: Can Rolle's theorem be used for (g(x)=\sin x) on ([0,\pi])?

Solution: 1. Continuity: (\sin x) is continuous for all real numbers → OK. 2. Differentiability: (\sin x) is differentiable everywhere → OK.
3. Endpoint values:
(g(0)=\sin 0 = 0)
(g(\pi)=\sin \pi = 0)
Hence (g(0)=g(\pi)) → OK.

Conclusion: All three conditions hold; Rolle's theorem guarantees at least one (c\in(0,\pi)) with (g'(c)=\cos c=0). Indeed, (c=\pi/2) works.

Example 3: Piecewise Function with a Corner

Problem: Examine (h(x)=|x|) on ([-1,1]).

Solution:

1. Continuity: The absolute value function is continuous everywhere → OK. 2. Differentiability: (h(x)) has a corner at (x=0); the left‑hand derivative is (-1), the right‑hand derivative is (+1). The derivative does not exist at (x=0), which lies inside ((-1,1)). → Differentiability fails.
2. Endpoint values: (h(-1)=|-1|=1), (h(1)=|1|=1) → equal

Extendingthe Checklist

When the three prerequisite tests are satisfied, the theorem does more than merely promise the existence of a stationary point; it also guarantees that the stationary point can be located somewhere inside the open interval. In practice, the search for that point often proceeds by solving the equation (f'(x)=0) after the initial verification. If the derivative is easy to factor or simplify, the algebraic solution may be immediate; otherwise, numerical techniques such as Newton’s method or bisection can be employed to locate a root of (f') within ((a,b)).

Example 4 – A Function with a Removable Discontinuity at an Endpoint

Consider
[ p(x)=\begin{cases} x^{2}, & -1\le x<0,\[4pt] 2x, & 0\le x\le 1 . \end{cases} ]

1. Continuity on ([-1,1]).
   The left‑hand limit at (0) equals (0), the right‑hand limit also equals (0), and (p(0)=0). Hence the function is continuous on the whole closed interval.

2. Differentiability on ((-1,1)).
   On ((-1,0)) the derivative is (2x); on ((0,1)) it is (2). Both formulas are defined everywhere inside the open interval, so differentiability holds there.

3. Endpoint values.
   (p(-1)=(-1)^{2}=1) and (p(1)=2\cdot1=2). Because the endpoint values differ, the third condition fails and Rolle’s theorem cannot be invoked.

Take‑away: Even when the interior derivative is well‑behaved, a mismatch at the boundaries disqualifies the theorem.

Example 5 – A Function That Meets All Three Criteria Yet Yields Multiple Critical Points

Let
[ q(x)=x^{4}-2x^{2}+1 \quad\text{on}\quad[-1,1]. ]

• Continuity & differentiability: As a polynomial, (q) is smooth on the entire real line.
• Endpoint equality: (q(-1)=(-1)^{4}-2(-1)^{2}+1=1-2+1=0) and (q(1)=1-2+1=0).

Since all hypotheses are satisfied, Rolle’s theorem assures at least one (c\in(-1,1)) with (q'(c)=0). Computing the derivative, [ q'(x)=4x^{3}-4x=4x(x^{2}-1), ] we obtain the stationary points (x=-1,0,1). Only (x=0) lies strictly inside the interval, confirming the theorem’s prediction and illustrating that more than one critical point may exist.

Example 6 – A Function That Passes the First Two Tests but Fails the Equality Test, Yet Still Possesses a Stationary Point

Take
[ r(x)=\cos x\quad\text{on}\quad[0,2\pi]. ]

• Continuity & differentiability: (\cos x) is continuous and infinitely differentiable everywhere.
• Endpoint values: