Why Check the Hypotheses First?
The Mean Value Theorem guarantees a point where the instantaneous rate of change equals the average rate of change—but only when its hypotheses hold. Before using that conclusion, you must check the function on the entire interval, not just at a few convenient points.
The check has two distinct parts. The function must be continuous on the closed interval, including its endpoints, and differentiable at every point in the open interval. Piecewise formulas require special attention wherever the rule changes: smoothness of each separate formula does not automatically make the joined function differentiable.
What Must Be Checked?
Continuous on \([a,b]\) means continuous at every point between \(a\) and \(b\), with the appropriate one-sided continuity at the endpoints. In particular, at \(a\) the function must approach \(f(a)\) from within the interval, and at \(b\) it must approach \(f(b)\) from within the interval.
Differentiable on \((a,b)\) means that the ordinary derivative exists at every interior point. The endpoints are not included in this differentiability condition. A function may satisfy the theorem’s hypotheses even if an endpoint derivative does not exist.
For a formula given by familiar functions, check that it is defined throughout the interval and that its components are continuous there. Then check that it is differentiable at every interior point. For a piecewise formula, also check continuity and differentiability at each interior joining point.
Identify \(a\) and \(b\), verify \(a<b\), and keep the closed interval and open interval conditions distinct.
Check the whole interval, including endpoint behavior and any piecewise joins.
Check every interior point, paying particular attention to corners, cusps, vertical tangents, and formula changes.
If both conditions hold, the theorem guarantees at least one \(c\) in the open interval. It does not place \(c\) at an endpoint.
Piecewise Formulas at a Joining Point
Suppose a piecewise function changes formula at an interior point \(d\). Each branch might be differentiable on its own side of \(d\), but the full function is differentiable at \(d\) only if the difference quotient has the same limit from the left and the right.
If the branches extend smoothly to \(d\), the practical test is: first verify that their values agree with the defined value \(f(d)\); then compare their derivatives at \(d\). More precisely, the left and right difference-quotient limits must exist and be equal. When they are equal, the derivative at \(d\) exists and equals that common value.
Why Matching One-Sided Derivatives Works
This test follows directly from the definition of a derivative. The derivative at \(d\) exists if the two-sided limit of \(\frac{f(x)-f(d)}{x-d}\) exists as a finite number. As \(x\) approaches \(d\), it does so either from the left or from the right. Thus, if the two one-sided limits exist and have the same value \(L\), the two-sided limit exists and equals \(L\). Conversely, if the two-sided limit exists, both one-sided limits must exist and equal that same limit.
Continuity is part of the check because the difference quotient uses the actual value \(f(d)\). For smooth branches meeting at \(d\), matching branch values establishes continuity; matching their derivatives then establishes differentiability. A mismatch in either check prevents the corresponding conclusion.
Worked Examples
Worked Example: A Square-Root Function at an Endpoint
Check whether \(f(x)=\sqrt{x}\) satisfies the Mean Value Theorem hypotheses on \([0,4]\).
Continuity: The square-root function is defined and continuous for \(x\geq 0\), so it is continuous on the entire closed interval \([0,4]\). At the endpoints, \(f(0)=0\) and \(f(4)=2\); the function approaches these values from within the interval.
Differentiability: For every interior point \(0<x<4\), the derivative exists and is \(f'(x)=\frac{1}{2\sqrt{x}}\). This expression is defined for every \(x\in(0,4)\).
Although the derivative formula is not defined at \(x=0\), that does not violate the theorem’s differentiability condition: \(0\) is an endpoint, not an interior point. The function is continuous on \([0,4]\) and differentiable on \((0,4)\), so the theorem applies. As a check on the secant slope, \(\frac{f(4)-f(0)}{4-0}=\frac{2-0}{4}= \frac12\). This calculation does not find a mean value point; it identifies the slope that the theorem says \(f'\) attains somewhere inside the interval.
Worked Example: A Piecewise Function with a Smooth Join
Check the hypotheses for \(f(x)=\begin{cases}x^2,&x\leq 1,\\2x-1,&x>1,\end{cases}\) on \([0,2]\).
Each branch is defined and continuous on its own part of the interval. At the join \(x=1\), the left branch gives \(1^2=1\), and the defined value is \(f(1)=1\). The right branch approaches \(2(1)-1=1\). Therefore, the function is continuous at \(1\), and it is continuous throughout \([0,2]\). The endpoints also have finite values: \(f(0)=0\) and \(f(2)=3\).
On the open parts of the interval, the derivatives are \(2x\) for \(x<1\) and \(2\) for \(x>1\). At the join, calculate directly: \(\lim_{x\to1^-}\frac{f(x)-f(1)}{x-1} =\lim_{x\to1^-}\frac{x^2-1}{x-1} =\lim_{x\to1^-}(x+1)=2\). From the right, \(\lim_{x\to1^+}\frac{f(x)-f(1)}{x-1} =\lim_{x\to1^+}\frac{(2x-1)-1}{x-1} =\lim_{x\to1^+}\frac{2x-2}{x-1}=2\). The one-sided limits agree, so \(f'(1)=2\). Thus \(f\) is differentiable at the only interior join, and it is differentiable elsewhere in \((0,2)\). Both hypotheses hold, so the Mean Value Theorem applies. As a numerical check, the secant slope is \(\frac{f(2)-f(0)}{2-0}=\frac{3-0}{2}=\frac32\).
Worked Example: A Continuous Function with a Corner
Check whether \(g(x)=|x|\) satisfies the theorem’s hypotheses on \([-1,1]\).
The absolute-value function is continuous for all real \(x\). In particular, \(g(-1)=1\), \(g(0)=0\), and \(g(1)=1\), and the one-sided limits at \(0\) both equal \(0=g(0)\). So \(g\) is continuous on \([-1,1]\).
For \(x<0\), \(g(x)=-x\), and for \(x>0\), \(g(x)=x\). At the interior point \(0\), the left difference quotient is \(\frac{g(x)-g(0)}{x-0}=\frac{-x}{x}=-1\) for \(x<0\), so its limit is \(-1\). The right difference quotient is \(\frac{x}{x}=1\) for \(x>0\), so its limit is \(1\). Since these limits differ, \(g'(0)\) does not exist.
The function is not differentiable on the entire open interval \((-1,1)\). Therefore, the Mean Value Theorem hypotheses are not satisfied, and its conclusion cannot be invoked on this interval.
A Common Mistake
A tempting but incorrect argument for the preceding piecewise example is: “The left branch \(-x\) and the right branch \(x\) are both differentiable, so \(|x|\) is differentiable at \(0\).” This checks each formula separately but skips the point where the formulas meet.
The correction is to test the derivative at the join itself. For \(|x|\), the left-hand derivative is \(-1\), while the right-hand derivative is \(1\). Since \(-1\ne 1\), the derivative at \(0\) does not exist. Smooth pieces do not guarantee a smooth join.
Check Your Understanding
For each question, distinguish carefully between what must be checked at endpoints and what must be checked in the interior.
- State the two hypotheses required by the Mean Value Theorem on \([a,b]\).
- Does failure of a derivative to exist at \(a\) alone prevent the theorem from applying on \([a,b]\)? Explain.
- For a piecewise function with an interior join at \(d\), what must be checked to establish continuity at \(d\)?
- What two limits determine whether the derivative exists at an interior join?
- Why is it not enough to know that each branch of a piecewise formula is differentiable on its own interval?
