Why Check the Hypotheses First?
Rolle’s Theorem is useful only when its hypotheses are satisfied. Before using it for a function on an interval, you must verify three separate facts: the function is continuous on the entire closed interval, differentiable at every point strictly inside it, and has equal values at the two endpoints.
These checks answer different questions. Continuity rules out breaks on the interval, differentiability rules out interior corners or other failures of a derivative, and endpoint equality is a direct comparison of two function values. Establishing two conditions does not make the third automatic.
Continuity on \([a,b]\) includes the endpoints. At \(a\), the relevant limit is from within the interval, so \(f(x)\) must approach \(f(a)\) as \(x\) approaches \(a\) from the right. At \(b\), it must approach \(f(b)\) from the left. At each interior point, continuity has its usual two-sided meaning.
Differentiability is required at every point in the open interval \((a,b)\), but not at \(a\) or \(b\). This distinction matters: a formula can have a vertical tangent or another derivative failure at an endpoint and still satisfy the differentiability hypothesis, provided it is differentiable at every interior point.
A Reliable Checking Process
For a particular function and interval, keep the checks separate. First make sure the function is defined throughout the closed interval. Then check continuity, including any piecewise joins and the endpoints. Next check differentiability at every interior point, paying special attention to corners, cusps, or points where a formula changes. Finally evaluate the function at the two endpoints and compare the results.
Confirm that \(f(x)\) has a defined value for every \(x\in[a,b]\).
Examine the whole interval, including one-sided continuity at its endpoints and both sides of any interior joins.
Verify differentiability at every \(x\in(a,b)\); do not require derivatives at \(a\) or \(b\).
Calculate \(f(a)\) and \(f(b)\), then determine whether they are equal.
For familiar functions, standard rules make some checks quick: polynomials are continuous and differentiable everywhere, and compositions of differentiable functions are differentiable wherever the composition is defined. Such facts help with routine intervals, but they do not remove the need to check the interval and any exceptional points in a piecewise or restricted function.
Worked Example: A Polynomial with Equal Endpoint Values
Worked Example: A Polynomial with Equal Endpoint Values
Check the hypotheses for \(f(x)=x^2-4x+3\) on \([1,3]\).
Continuity: The function is a polynomial, so it is continuous for every real \(x\). In particular, it is defined and continuous on the entire closed interval \([1,3]\).
Differentiability: A polynomial is differentiable everywhere. Its derivative is \(f'(x)=2x-4\), so \(f\) is differentiable at every point of the required open interval \((1,3)\).
Endpoint values:
Thus \(f(1)=f(3)\). All three hypotheses hold, so Rolle’s Theorem is applicable on \([1,3]\). As a check on the arithmetic, substituting \(1\) and \(3\) into the original formula gives \(0\) in both cases.
This example illustrates why the endpoint comparison should be written out. A function may be continuous and differentiable on the interval, but endpoint equality is a separate requirement; here it holds because the two explicit evaluations agree.
Checking a Nondifferentiable Interior Point
An expression can be continuous everywhere on an interval and still fail the differentiability condition. Absolute value provides a simple example: its graph has a corner at zero. To confirm that this is a genuine derivative failure, use the definition of derivative rather than relying only on the graph’s appearance.
Worked Example: Equal Endpoint Values but a Corner
Check \(f(x)=|x|\) on \([-1,1]\).
Continuity: The absolute-value function is continuous for every real \(x\), so it is continuous on \([-1,1]\).
Endpoint values: Since \(|-1|=1\) and \(|1|=1\), we have \(f(-1)=f(1)=1\).
Differentiability: The point \(0\) lies in \((-1,1)\). At this point, the difference quotient is
For \(h>0\), this quotient equals \(1\). For \(h<0\), it equals \(-1\). Therefore its two-sided limit as \(h\) approaches zero does not exist, so \(f'(0)\) does not exist. The function is not differentiable at every point of \((-1,1)\), and the hypotheses fail.
As a check, the endpoint values really are equal, and continuity holds; the failure is specifically the interior differentiability condition at \(x=0\).
The difference quotient calculation also explains the general issue: to have a derivative at an interior point, the same finite limit must result from approaching from either side. Here the left-hand and right-hand limits disagree.
Checking Continuity at a Piecewise Join
For a piecewise function, it is not enough to observe that each formula is continuous on the part where it is used. At a join, the values and limits from the pieces must fit together. If the one-sided limits at an interior join disagree, the function is not continuous there.
Worked Example: Endpoint Equality Does Not Repair a Break
Define \(f\) on \([-1,1]\) by
Domain and endpoint values: The definition assigns a value to every point of \([-1,1]\). At the endpoints,
Thus \(f(-1)=f(1)\).
Continuity at the join: As \(x\) approaches \(0\) from the left, the formula \(x+1\) gives a limit of \(1\). From the right, the formula \(x-1\) gives a limit of \(-1\). Since these one-sided limits differ, the two-sided limit at \(0\) does not exist. In particular, it cannot equal \(f(0)=0\), so \(f\) is not continuous on \([-1,1]\).
The function is not differentiable at \(0\) either, but the continuity failure alone is enough to rule out applying Rolle’s Theorem. The endpoint equality does not compensate for the break in the interior.
Endpoint Differentiability Is Not Required
A frequent over-check is to demand differentiability at \(a\) and \(b\). The condition is differentiability on the open interval only. The next example satisfies the hypotheses even though the displayed derivative formula has a zero denominator at both endpoints.
Worked Example: Derivative Failures at the Endpoints
Check \(f(x)=\sqrt{x(1-x)}\) on \([0,1]\).
Continuity: For \(0\le x\le1\), both \(x\) and \(1-x\) are nonnegative, so \(x(1-x)\ge0\). The product \(x(1-x)\) is continuous, and the square-root function is continuous on the nonnegative real numbers. Therefore their composition \(f\) is continuous on all of \([0,1]\).
Differentiability in the interior: If \(0<x<1\), then \(x(1-x)>0\), so the square root is differentiable there. By the chain rule,
The denominator is positive for every \(x\in(0,1)\), so this derivative exists at every interior point.
Endpoint values:
All three hypotheses hold. The derivative expression is not defined at \(x=0\) or \(x=1\), but those points are endpoints, not members of the required differentiability interval \((0,1)\). The continuity check at the endpoints still matters and has already been established.
When Endpoint Equality Fails
Regularity does not imply equal endpoint values. A simple function can be continuous on the closed interval and differentiable throughout its interior while having different endpoint values.
Worked Example: A Smooth Function with Unequal Endpoints
Check \(f(x)=x\) on \([0,1]\).
The function is a polynomial, hence continuous on \([0,1]\) and differentiable on \((0,1)\), with derivative \(f'(x)=1\). But
Since \(0\ne1\), the endpoint-equality hypothesis fails. Thus the other two conditions do not suffice to apply Rolle’s Theorem. Substituting the endpoint values into the original function confirms the comparison.
Common Mistakes and a Final Checklist
A common mistake is to check only that a formula looks smooth and that the endpoints match. For example, with \(f(x)=|x|\) on \([-1,1]\), one might write “continuous, endpoints both equal \(1\), so the hypotheses hold.” That conclusion is wrong: it omits the differentiability check at \(0\), where the left and right difference quotients are \(-1\) and \(1\). The correction is to test every interior point where the formula or graph may fail to be differentiable.
Another mistake is to require \(f'(a)\) and \(f'(b)\). That is not part of the condition. The function must be continuous on the closed interval and differentiable in its interior; no endpoint derivative is demanded.
Check Your Understanding
For each question, identify which hypothesis is being tested and explain the relevant check.
- For \(f(x)=x^2-1\) on \([-1,1]\), verify continuity, interior differentiability, and endpoint equality.
- For \(f(x)=|x-2|\) on \([1,3]\), which hypothesis fails, and how can the failure be shown using a difference quotient?
- Why must continuity be checked at the endpoints even though differentiability is required only on the open interval?
- Can a function satisfy the differentiability condition if its derivative is undefined at one endpoint? Explain what interval the condition concerns.
- For \(f(x)=x+2\) on \([0,1]\), which of the three hypotheses fails?
