Mean Value Theorem
The mean value theorem states that if f is continuous on [a, b] and differentiable on (a, b), then there is a point c in (a, b) where the instantaneous rate of change equals the average rate of change: f′(c) = [f(b) − f(a)]/(b − a). In brief: if a car travels 120 km in 2 hours, at some moment its speed is exactly 60 km/h. For f(x) = x2 on [1, 3], the average slope is (9 − 1)/2 = 4 and f′(c) = 2c = 4 gives c = 2. Rolle’s theorem is the special case f(a) = f(b), where f′(c) = 0.
The mean value theorem (MVT) links the derivative at a point with the behaviour of the function on an interval. It is a theoretical tool with many practical consequences: a function with zero derivative is constant, a function with a positive derivative is increasing, and the error of a linear approximation can be bounded. This guide states Rolle’s theorem and the MVT with a proof idea, shows how to find the point c, derives the main consequences, covers the Cauchy form used in l’Hôpital’s rule and gives applications and a table of examples.
What is Rolle’s theorem?
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is a c in (a, b) with f′(c) = 0. Reason: by the extreme value theorem, f has a maximum and a minimum on [a, b]. If both are at the endpoints, f is constant and f′ = 0 everywhere. Otherwise, there is an extremum at an interior point c, and by Fermat’s theorem, f′(c) = 0. The graph has a horizontal tangent somewhere between two points of equal height.
Example 1
f(x) = x2 − 4x + 3 on [1, 3]: f(1) = 0 = f(3). f′(x) = 2x − 4 = 0 at x = 2, which lies in (1, 3) ✓.
What does the mean value theorem say?
If f is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) such that f′(c) = (f(b) − f(a))/(b − a). Geometrically: the tangent at c is parallel to the secant line through (a, f(a)) and (b, f(b)).
Proof idea: subtract from f the linear function whose graph is the secant line. The difference h(x) = f(x) − f(a) − [(f(b) − f(a))/(b − a)](x − a) satisfies h(a) = h(b) = 0, so by Rolle’s theorem h′(c) = 0, which gives the claim.
How do you find the point c?
- Check the hypotheses: continuity on [a, b] and differentiability on (a, b).
- Compute the average rate m = (f(b) − f(a))/(b − a).
- Solve f′(c) = m and keep the solutions inside (a, b).
Example 2
f(x) = x3 on [0, 2]: m = (8 − 0)/2 = 4, and 3c2 = 4 gives c = 2/√3 ≈ 1.155 (the negative root is not in the interval).
Example 3
f(x) = √x on [1, 9]: m = (3 − 1)/8 = 1/4. f′(c) = 1/(2√c) = 1/4 gives √c = 2, c = 4.
Example 4
f(x) = ln x on [1, e]: m = 1/(e − 1) ≈ 0.582, and 1/c = 0.582 gives c = e − 1 ≈ 1.718, in (1, e) ✓.
Example 5: when the hypotheses fail
f(x) = |x| on [−1, 2]: m = (2 − 1)/3 = 1/3. But f′ = ±1 and never equals 1/3 (and f is not differentiable at 0), so no c exists. For f(x) = 1/x on [−1, 1], f is not continuous at 0, and the average slope 1 would require f′(c) = −1/c2 = 1, which has no solution (f′ is always negative). The theorem does not apply.
What are the consequences?
- Zero derivative: if f′ = 0 on an interval, f is constant there. Two functions with the same derivative differ by a constant.
- Monotonicity: if f′ > 0 on an interval, f is increasing, and if f′ < 0, decreasing. Proof: for x1 < x2, f(x2) − f(x1) = f′(c)(x2 − x1) has the sign of f′(c).
- Bounded derivative implies Lipschitz: if |f′| ≤ M, then |f(x) − f(y)| ≤ M|x − y|.
- Uniqueness of roots: if f′ ≠ 0 on an interval, f has at most one zero there (otherwise Rolle gives a zero of f′).
- Error of linear approximation (Taylor): f(x) − f(a) = f′(c)(x − a) for some c between a and x.
What are the applications?
Example 6: an inequality
Show that |sin x − sin y| ≤ |x − y|. By the MVT, sin x − sin y = cos c · (x − y) for some c, and |cos c| ≤ 1 ✓.
Example 7: the inequality ex ≥ 1 + x
For x > 0: ex − 1 = ec · x with c between 0 and x, so c > 0 and ec > 1, giving ex − 1 > x. Similarly for x < 0.
Example 8: a bound for a root
Estimate √26 − 5. With f(x) = √x on [25, 26]: f(26) − f(25) = f′(c) · 1 = 1/(2√c) for some c in (25, 26). Since 5 < √c < √26 < 5.1, we get 1/10.2 < √26 − 5 < 1/10, i.e., between 0.0980 and 0.1. The true value is 0.0990.
Example 9: a speed check
A car passes a toll gate at 10:00 and another gate 150 km away at 11:30. The average speed is 100 km/h. By the MVT, at some moment, the car drove exactly 100 km/h, and if the speed limit is 90 km/h, the car must have been speeding at some point.
Example 10: counting solutions
Show that x3 + 3x − 5 = 0 has exactly one real root. f(1) = −1 < 0 and f(2) = 9 > 0, so a root exists in (1, 2) by the intermediate value theorem. If there were two roots, Rolle’s theorem would give a point with f′ = 3x2 + 3 = 0, which is impossible. So the root is unique.
Example 11: the Cauchy form
For two functions f and g satisfying the conditions, with g′ ≠ 0: [f(b) − f(a)]/[g(b) − g(a)] = f′(c)/g′(c). This generalisation is used in the proof of l’Hôpital’s rule.
Table of examples
| Function | Interval | Average slope | Point c |
|---|---|---|---|
| x2 | [1, 3] | 4 | 2 |
| x3 | [0, 2] | 4 | 2/√3 ≈ 1.155 |
| √x | [1, 9] | 1/4 | 4 |
| ln x | [1, e] | 1/(e − 1) | e − 1 ≈ 1.718 |
| x2 − 4x + 3 (Rolle) | [1, 3] | 0 | 2 |
| |x| | [−1, 2] | 1/3 | none (not differentiable at 0) |
Where does the theorem come from?
Michel Rolle proved a special case in 1691 for polynomials, in the context of locating roots, and was in fact a critic of calculus at the time. The mean value theorem in its modern form was formulated by Joseph-Louis Lagrange in the 1770s, and Cauchy gave the rigorous proof in 1823 with the generalisation named after him. The name “Rolle’s theorem” was given later, in the nineteenth century.
Where is it used?
The MVT justifies the use of the derivative sign to determine the monotonicity. It provides error bounds in numerical analysis, where it estimates how much a function can change, and it is the key to the fundamental theorem of calculus. In real life, it underlies statements such as “average speed cameras can detect speeding between two points”.
Which mistakes are common?
Forgetting to check the hypotheses
Continuity on the closed interval and differentiability on the open interval are required. |x| on [−1, 1] satisfies Rolle’s endpoint condition, but no c exists.
Assuming that c is the midpoint
c is not necessarily (a + b)/2. For x3 on [0, 2], c ≈ 1.155, while the midpoint is 1.
Giving a c outside the interval
Solve f′(c) = m and keep only solutions in (a, b).
Assuming c is unique
There may be several points. For sin x on [0, 2π] (Rolle), f′ = cos x = 0 at both π/2 and 3π/2.