Why Check the Hypotheses First?
The Mean Value Theorem guarantees a point where a function’s instantaneous rate of change equals its average rate of change. That guarantee is conditional: the function must be continuous across the whole closed interval and differentiable throughout its interior. Before using the theorem, we therefore need to check those conditions for the function as it is actually defined on the specified interval.
The check is usually straightforward for a polynomial. It requires more care for a rational function, whose denominator might vanish, or for a piecewise function, whose formula might change at an interior point. Endpoint behavior also matters: the function must be defined and continuous at both endpoints, even though it does not need to be differentiable there.
The Conditions to Check
These hypotheses ask for two different checks on two different sets:
- On the closed interval \([a,b]\): \(f\) must be defined and continuous, including at \(a\) and \(b\).
- On the open interval \((a,b)\): \(f\) must be differentiable at every point, including any interior point where a piecewise formula changes.
Continuity at an interior point \(x_0\) means that \(f(x_0)\) is defined and \(\lim_{x\to x_0}f(x)=f(x_0)\). At the endpoints, continuity is one-sided: the limit from within the interval must equal the endpoint value. Differentiability at an interior point means that the finite limit \(\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}\) exists. The Mean Value Theorem does not require endpoint derivatives.
When both conditions hold, the theorem guarantees a suitable \(c\). Checking the hypotheses does not itself identify that point; it tells us whether the guarantee is available.
A Practical Checklist
Confirm that the endpoints satisfy \(a<b\), and note every formula change or possible domain restriction between them.
Verify that the function is defined and continuous at every point of \([a,b]\). Pay particular attention to the endpoints and to zeros of a rational denominator.
Verify differentiability at every point of \((a,b)\). For a piecewise function, check each interior junction as well as the smoothness of each formula between junctions.
If both hypotheses hold, the theorem guarantees at least one \(c\in(a,b)\). If a hypothesis fails, the theorem does not apply on that interval.
Worked Example: A Polynomial on a Closed Interval
Worked Example: A Polynomial on a Closed Interval
Determine whether \(f(x)=2x^3-3x+1\) satisfies the Mean Value Theorem’s hypotheses on \([-1,2]\).
The endpoints are \(a=-1\) and \(b=2\), so \(a<b\). A polynomial is defined and continuous for every real number, and its derivative is also a polynomial. In particular, \(f\) is continuous on \([-1,2]\) and differentiable on \((-1,2)\). Both hypotheses hold.
For completeness, the endpoint values and the average rate of change are:
The theorem therefore guarantees at least one \(c\in(-1,2)\) with \(f'(c)=3\). This is a conclusion from the verified hypotheses; no specific value of \(c\) is needed to decide whether the theorem applies.
Check: The endpoint difference is \(11-2=9\), and the interval length is \(2-(-1)=3\), so the computed average rate is \(9/3=3\).
Rational Functions: Check the Original Domain
A rational function is a quotient of polynomials. Wherever its denominator is nonzero, it is continuous and differentiable. Thus, for a rational formula, a useful first test is whether the denominator vanishes anywhere on the specified closed interval. A zero in the interval means the original formula is not defined there, unless the function has separately been defined at that point.
Worked Example: A Rational Function with No Denominator Zeros
Determine whether \(f(x)=\frac{x+1}{x^2+1}\) satisfies the Mean Value Theorem’s hypotheses on \([-1,1]\).
For every real \(x\), \(x^2\geq 0\), so \(x^2+1\geq 1>0\). The denominator is never zero. The quotient is therefore defined, continuous, and differentiable throughout \([-1,1]\), including at the endpoints. The hypotheses hold.
We can also calculate the endpoint values and average rate:
The Mean Value Theorem guarantees at least one \(c\in(-1,1)\) with \(f'(c)=\frac12\). The derivative exists throughout the interval; for instance, applying the quotient rule gives
Check: Substitution into the original quotient gives \(f(-1)=0\) and \(f(1)=1\); their difference is \(1\), and the interval length is \(2\), confirming the average rate \(\frac12\).
Worked Example: A Removable-Looking Hole
Worked Example: A Removable-Looking Hole
Consider \(f(x)=\frac{x^2-1}{x-1}\) on \([0,2]\), with the function defined by this quotient. Does it satisfy the Mean Value Theorem’s hypotheses?
At \(x=1\), the denominator is \(1-1=0\), while the numerator is \(1^2-1=0\). The quotient is therefore undefined at \(x=1\), which lies inside \([0,2]\). Consequently, the function as given is not defined or continuous on the whole closed interval, so the Mean Value Theorem’s hypotheses fail.
For \(x\neq 1\), factoring and cancelling gives
The cancellation is valid only when \(x\neq 1\). It does not assign a value to the original quotient at \(x=1\). If instead we defined a new function by setting \(f(1)=2\), then the resulting function would be \(x+1\) on \([0,2]\), and it would satisfy the hypotheses. That extension is a different definition from the original quotient.
Check: At \(x=1\), direct substitution into the original formula gives \(\frac{0}{0}\), which is undefined. The simplified expression has value \(2\) there, but that does not change the domain of the original function.
Piecewise Functions: Inspect Every Interior Junction
For a piecewise function, each formula may be differentiable on its own part of the interval, but that alone is not enough. At an interior junction, the pieces must fit together continuously, and the function must also be differentiable there. Matching function values or limits checks continuity; it does not by itself establish differentiability.
The derivative test follows directly from the definition. For \(h<0\), the difference quotient uses points to the left of \(x_0\); for \(h>0\), it uses points to the right. The two-sided limit as \(h\to 0\) exists precisely when these one-sided limits exist and agree. This explains why a piecewise function’s interior junction must be checked from both sides.
Worked Example: A Piecewise Function That Joins Differentiably
Determine whether the function
satisfies the Mean Value Theorem’s hypotheses on \([0,2]\).
On the left side of the junction, \(f(1)=1^2=1\). The right-hand limit is \(\lim_{x\to 1^+}(2x-1)=2(1)-1=1\). The left-hand limit is \(\lim_{x\to 1^-}x^2=1^2=1\). Both one-sided limits equal the defined value \(f(1)=1\), so \(f\) is continuous at the junction.
To check differentiability there, calculate the difference quotient from each side. If \(h<0\), then \(1+h\leq 1\), and
Its limit as \(h\to0^-\) is \(2\). If \(h>0\), then \(1+h>1\), and
Its limit as \(h\to0^+\) is also \(2\). The two one-sided limits agree, so \(f'(1)=2\). Away from \(x=1\), each piece is a polynomial and hence differentiable. The function is continuous at the endpoints \(0\) and \(2\) as well, since the relevant formulas are polynomials there. Therefore, \(f\) is continuous on \([0,2]\) and differentiable on \((0,2)\); the theorem applies.
Check: The left and right difference-quotient limits are both \(2\), confirming the derivative at the only interior junction. The endpoints are included in the continuity check, not the differentiability check.
Endpoint Checks and Common Errors
At an endpoint, continuity must be checked using the values and limits from within the interval. For example, if a piecewise definition assigns \(f(0)=1\) but uses \(f(x)=x\) for \(0<x\leq 1\), then \(\lim_{x\to0^+}f(x)=0\neq f(0)\). The function is not continuous on \([0,1]\), so the theorem does not apply there. A derivative at \(0\) is not required; the failed endpoint continuity is enough.
A frequent error is to check only that a formula looks smooth between the endpoints. A rational formula can have a hole inside the interval, and a piecewise formula can have a corner at a junction. Another error is to demand differentiability at the endpoints: that is stronger than the theorem requires. The correct checks are continuity on the entire closed interval and differentiability at every interior point.
Check Your Understanding
For each question, focus on whether the hypotheses hold on the stated interval; do not try to find a Mean Value point.
- Does \(f(x)=x^4-2x+3\) satisfy the hypotheses on \([-2,1]\)? State which facts about polynomials you use.
- For \(f(x)=\frac{1}{x-3}\) on \([0,2]\), what must be checked about the denominator, and does the theorem apply?
- Why does cancelling \(x-1\) in \(\frac{x^2-1}{x-1}\) not show that the original quotient is defined at \(x=1\)?
- For a piecewise function with a junction strictly inside the interval, what must be checked there for continuity and for differentiability?
- Does the Mean Value Theorem require the function to have a derivative at either endpoint? Which condition does it require there instead?
