Rolle's theorem and Lagrange's Mean Value Theorem (MVT) are existence theorems: they tell us a point with a certain property must exist, without telling us where. Despite the modest statements, they are the engine of nearly every proof in single-variable calculus , monotonicity, Taylor's theorem, L'Hôpital's rule.
Rolle's theorem
Theorem. If f is
continuous on [a,b],
differentiable on (a,b), and
f(a)=f(b),
then there exists at least one c∈(a,b) such that f′(c)=0.
Geometric meaning. If a smooth curve starts and ends at the same height, it must have a horizontal tangent somewhere in between.
Lagrange's Mean Value Theorem
Theorem. If f is
continuous on [a,b] and
differentiable on (a,b),
then there exists at least one c∈(a,b) such that
f′(c)=b−af(b)−f(a).
Geometric meaning. The slope of the chord from (a,f(a)) to (b,f(b)) equals the slope of the tangent at some interior point c.
Proof of Lagrange's MVT from Rolle
Define g(x)=f(x)−L(x), where L(x) is the line through (a,f(a)) and (b,f(b)). Then g(a)=g(b)=0, so by Rolle's theorem there is c with g′(c)=0, i.e., f′(c)=L′(x)=b−af(b)−f(a). ■
Why the hypotheses matter
Rolle's theorem can fail if:
f is not continuous on [a,b]: take f with a jump.
f is not differentiable on (a,b): e.g., f(x)=∣x∣ on [−1,1].
f(a)=f(b).
Similar comments for Lagrange.
Consequences
Monotonicity: If f′(x)>0 on an interval, then f is strictly increasing on it. Reason: for any a<b in the interval, MVT gives f(b)−f(a)=f′(c)(b−a)>0.
Constants: If f′(x)=0 on an interval, then f is constant.
Bounded slopes: If ∣f′(x)∣≤M on an interval, then ∣f(b)−f(a)∣≤M∣b−a∣ , Lipschitz condition.
Worked example
Verify Rolle's theorem for f(x)=x2−4x+3 on [1,3].
f(1)=0,f(3)=0. Continuous and differentiable everywhere. By Rolle, ∃c∈(1,3) with f′(c)=0. f′(c)=2c−4=0⇒c=2. ✓
Verifying Lagrange's MVT
For f(x)=x3 on [1,4]:
Slope of chord = 364−1=21. MVT promises c∈(1,4) with f′(c)=21, i.e., 3c2=21, so c=7≈2.65∈(1,4). ✓