Why Test What Happens When a Hypothesis Fails?
Rolle’s Theorem and the Mean Value Theorem guarantee that a derivative takes a specified value somewhere inside an interval. Their hypotheses are essential to those guarantees: continuity on the closed interval and differentiability on its interior are what allow the conclusions to be established.
A useful way to understand a theorem’s conditions is to remove one condition and look for a function whose conclusion fails. Such a function is a counterexample to the claim that the conclusion follows without that condition. It does not show that the theorem is false; it shows why the theorem cannot be applied when a required hypothesis is missing.
The Guarantees Being Tested
Rolle’s Theorem states that if \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b)\), then there is some \(c\in(a,b)\) such that \(f'(c)=0\).
The Mean Value Theorem states that if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is some \(c\in(a,b)\) such that
For the Mean Value Theorem, the right-hand side is the slope of the secant line through the endpoint values. For Rolle’s Theorem, the endpoint values agree, so that secant slope is zero. The examples below check whether the conclusion actually occurs after one of the required conditions is removed.
Why the Hypotheses Enter the Argument
The link between the two theorems can be seen by subtracting the secant line from \(f\). Suppose first that \(f\) satisfies the Mean Value Theorem’s hypotheses, and define
The subtracted expression is the line through \((a,f(a))\) and \((b,f(b))\). Substitution gives \(g(a)=0\). Also,
Since \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), so is \(g\). Rolle’s Theorem therefore gives an interior point \(c\) with \(g'(c)=0\). Differentiating the definition of \(g\) yields \(g'(x)=f'(x)-m\), so \(f'(c)=m\). This derivation shows where the conditions are used: without continuity or differentiability, the Rolle step may not be available.
When an Endpoint Is Discontinuous
Worked Example: Equal Endpoint Values but No Zero Derivative
Define \(f\) on \([0,1]\) by
The endpoint values are \(f(0)=1\) and \(f(1)=1\), so the equal-endpoint condition in Rolle’s Theorem is satisfied. For every \(x\in(0,1)\), the function is \(f(x)=x\), and hence \(f'(x)=1\). There is no \(c\in(0,1)\) for which \(f'(c)=0\).
The missing hypothesis is continuity at the left endpoint. In fact, \(\lim_{x\to 0^+}f(x)=0\), whereas \(f(0)=1\). Thus \(f\) is not continuous on \([0,1]\). The equal endpoint values alone do not guarantee a point with zero derivative.
Check: the endpoint values are both \(1\), and the derivative at every interior point is \(1\), so the claimed Rolle conclusion really does fail.
Worked Example: The Mean Value Conclusion Fails at an Endpoint Discontinuity
Define \(f\) on \([0,1]\) by
The secant slope calculated from the assigned endpoint values is
For every \(x\in(0,1)\), \(f'(x)=1\). Therefore there is no interior point \(c\) with \(f'(c)=-1\), as the Mean Value Theorem’s conclusion would require. Here, too, continuity fails at \(0\): the right-hand limit is \(0\), not the assigned value \(2\).
Check: the secant slope is \(-1\), while every interior derivative is \(1\). The discrepancy is exactly the failure the theorem’s conclusion would rule out under its hypotheses.
When the Discontinuity Is Inside the Interval
Worked Example: An Interior Jump with the Wrong Secant Slope
Define \(f\) on \([0,1]\) by
The endpoint values are \(f(0)=0\) and \(f(1)=2\), so the secant slope is
On the open interval to the left of \(1/2\), \(f'(x)=1\). On the open interval to the right, differentiating \(x+1\) also gives \(f'(x)=1\). At \(x=1/2\), the left-hand limit is \(1/2\), but the function value and right-hand limit are \(3/2\). Thus \(f\) is discontinuous there, and its derivative is not defined there. At every point where the derivative exists in \((0,1)\), it equals \(1\), never the secant slope \(2\).
Check: the only interior point not covered by the two derivative calculations is the jump itself, where differentiability fails. Consequently, no interior point satisfies the Mean Value Theorem equation.
An interior jump can also defeat Rolle’s conclusion. For example, set \(q(x)=x\) for \(0\leq x<1/2\) and \(q(x)=x-1\) for \(1/2\leq x\leq1\). Then \(q(0)=q(1)=0\). Away from \(1/2\), \(q'(x)=1\), and at \(1/2\) the function is discontinuous. There is no interior point where the derivative is zero. The equal endpoint values do not compensate for the missing continuity.
When a Corner Prevents Differentiability
Worked Example: A Corner with Equal Endpoint Values
Consider \(f(x)=|x|\) on \([-1,1]\). This function is continuous, and its endpoint values agree:
For \(x<0\), \(|x|=-x\), so \(f'(x)=-1\). For \(x>0\), \(|x|=x\), so \(f'(x)=1\). At \(x=0\), the difference quotient is
The two one-sided limits differ, so \(f'(0)\) does not exist. At every point in \((-1,1)\) where \(f'\) exists, its value is either \(-1\) or \(1\), never \(0\). Rolle’s conclusion therefore fails because differentiability on the entire open interval is missing at the corner.
The Mean Value Theorem conclusion fails on this same interval as well. Its secant slope is \((1-1)/(1-(-1))=0\), but no differentiable interior point has derivative \(0\). The function is continuous on the closed interval; the absent condition is differentiability at the interior corner.
A Common Mistake: Treating a Missing Hypothesis as Automatic
A common error is to calculate the secant slope and then assume that some derivative must equal it, even when the function is discontinuous or has a corner. For the jump-function example, one might write, “The secant slope is \(2\), so \(f'(c)=2\) for some \(c\).” That last step is not a calculation: it is precisely the Mean Value Theorem’s conclusion, and the theorem cannot be applied because continuity fails at \(1/2\). Direct differentiation shows instead that \(f'(x)=1\) wherever the derivative exists in the interior.
A related mistake is to assign a derivative at a corner by choosing one of the one-sided slopes. For \(|x|\) at \(0\), the left-hand slope is \(-1\) and the right-hand slope is \(1\). Since these are unequal, there is no derivative at \(0\); neither value can be substituted as \(f'(0)\).
What Counterexamples Establish
The examples isolate three different failures: a discontinuity at an endpoint, a jump inside the interval, and a nondifferentiable interior corner. In each case, endpoint values and derivatives can be computed explicitly, and the required conclusion can be checked rather than guessed.
These examples do not say that a conclusion can never hold when a hypothesis fails. They show that the hypotheses cannot simply be discarded while keeping the guarantee. For instance, a function with a discontinuity can still have a zero derivative somewhere for other reasons. The logical point is that a theorem provides a dependable conclusion only when its stated conditions are met.
Check Your Understanding
For each question, identify the relevant hypothesis and explain what the function’s endpoint values or derivatives show.
- For the first example, which Rolle hypothesis fails, and what is \(f'(x)\) at every interior point?
- For the endpoint-discontinuous Mean Value example, calculate the secant slope and compare it with the derivative on \((0,1)\).
- In the interior-jump example, why can the point \(x=1/2\) not serve as the Mean Value Point?
- For \(f(x)=|x|\) on \([-1,1]\), what are the left- and right-hand difference-quotient limits at \(0\)?
- Does a failed hypothesis prove that a theorem’s conclusion is false for every such function? Explain the distinction.
