MAT148H1 Chap.8 The Mean Value Theorem and Its Consequences
The Mean Value Theorem and Its Consequences
What the theorem says
Unit 5 of the syllabus centres on the Mean Value Theorem. Take f continuous on the closed interval [a, b] and differentiable at every point of the open interval (a, b). Some point c strictly between a and b then satisfies f′(c) = (f(b) − f(a))/(b − a). Geometrically, the secant line joining (a, f(a)) to (b, f(b)) has a parallel tangent somewhere in between.
Physically, a car averaging 80 km/h over a trip must read exactly 80 on its speedometer at some instant. The theorem is not constructive: it guarantees that c exists but gives no method for locating it. Most uses never need the value of c, only the fact that it lies inside the interval.
Rolle's Theorem and the proof strategy
Rolle's Theorem is the case where the endpoint values agree.
Under the same two conditions, when both endpoint values are zero, the derivative vanishes at some interior point c. Its proof combines two earlier results. The Extreme Value Theorem gives a maximum and minimum on [a, b]; if f is not constant, one of them is attained at an interior point.
A local extremum at an interior point where f is differentiable forces a zero derivative, because the one sided difference quotients have opposite signs. The full Mean Value Theorem then follows by subtracting the secant line: F(x) = f(x) − ((f(b) − f(a))/(b − a))(x − a) has equal values at a and b, so Rolle applies to F, and F′(c) = 0 rearranges to the desired equation.
Expect to reproduce this tilting argument in a proof question.
Why the hypotheses matter
Each condition earns its place. The function |x| on [−1, 1] is continuous but has a corner at 0; its secant slope is 0 while its derivative is only ever ±1, so the conclusion fails. A function that jumps at an endpoint can also break the theorem even when it is differentiable inside.
On the other hand, differentiability at the endpoints is not required: √x on [0, 9] satisfies the hypotheses, and indeed 1/(2√c) = 3/9 at c = 9/4. When writing a solution, state both hypotheses explicitly before quoting the conclusion. A proof that quotes the slope equation without checking continuity and differentiability is incomplete, however correct the algebra that follows.
Counterexamples like |x| are also a favourite way to test whether a hypothesis is understood.
Consequences you will use repeatedly
Several standard facts are corollaries. A function with zero derivative on an interval is constant there, since every difference f(p) − f(a) equals 0 times (p − a). Two functions with equal derivatives differ by a constant.
A positive derivative forces a function to increase, and a negative derivative forces it to decrease. Rolle's Theorem bounds the number of roots: between two roots of a differentiable function lies a root of its derivative, so a function whose derivative never vanishes has at most one root. Pairing this with the Intermediate Value Theorem proves a root is unique.
Finally, a bound on f′ becomes a bound on function values: if |f′| ≤ M on an interval, then |f(x) − f(y)| ≤ M|x − y| there. That single line proves inequalities such as |sin x − sin y| ≤ |x − y| and gives quick numerical estimates for square roots and logarithms.
What this chapter covers
- 01
Statement of the Mean Value Theorem with exact hypotheses
- 02
Rolle's Theorem and interior extrema
- 03
Proof of the MVT by tilting to Rolle
- 04
Counterexamples when a hypothesis fails
- 05
Zero derivative implies constant
- 06
Counting roots with Rolle and the IVT
- 07
Inequalities from derivative bounds
Worked example · free
Estimating √101 with the Mean Value Theorem
- 1f(x) = √x is continuous on [100, 101] and differentiable on (100, 101) with f′(x) = 1/(2√x), so the MVT applies.
- 2Some c in (100, 101) satisfies √101 − √100 = f′(c)(101 − 100) = 1/(2√c).
- 1Upper bound: c > 100 gives √c > 10, so 1/(2√c) < 1/20 = 0.05 and √101 < 10.05.
- 2Lower bound: c < 101 < 121 gives √c < 11, so 1/(2√c) > 1/22 ≈ 0.0455 and √101 > 10.045.
Key terms
- Mean Value Theorem
- The Mean Value Theorem is the result that a function continuous on the closed interval [a, b] and differentiable at each interior point has some c inside where f′(c) equals the secant slope (f(b) − f(a))/(b − a).
- Rolle's Theorem
- Rolle's Theorem is the statement that a continuous function on a closed interval, differentiable inside it and vanishing at both ends, has a horizontal tangent, f′(c) = 0, at some interior point c.
- Local Maximum
- A local maximum is a point c with a neighbourhood on which f(x) ≤ f(c). When c is interior and f is differentiable there, the derivative at c must equal zero.
- Secant Slope
- The secant slope is the quantity (f(b) − f(a))/(b − a), the gradient of the straight line through two points on a graph. It measures average rate of change across the interval.
- Lipschitz Bound
- A Lipschitz bound is an inequality |f(x) − f(y)| ≤ M|x − y| holding for all x and y in an interval. The Mean Value Theorem supplies one whenever |f′| ≤ M there.
The Mean Value Theorem and Its Consequences FAQ
How do you prove an equation has exactly one real solution?
Split the claim in two. Existence comes from the Intermediate Value Theorem: find inputs where the function changes sign. Uniqueness comes from Rolle: two roots would force a zero of the derivative between them, so show the derivative never vanishes.
Does the Mean Value Theorem need the function to be differentiable at the endpoints?
No. Continuity is required on the closed interval, but differentiability only on the open interior. Square root on [0, 4] therefore qualifies even though its slope is undefined at zero, while |x| on [−1, 1] fails because of its interior corner.
Can the Mean Value Theorem tell me the value of c?
Not in general. The theorem guarantees existence only. For simple functions, such as a quadratic, you can solve f′(c) equal to the secant slope directly, but proofs and estimates usually rely solely on c lying strictly between the endpoints, which is all the bound requires.
How is the Mean Value Theorem used to prove inequalities?
Write the difference f(b) − f(a) as f′(c) times (b − a), then bound f′(c) using only a less than c less than b. Monotone derivatives make this easy, producing results such as ln(1 + x) less than x for positive x, or the estimate that √101 exceeds 10 by less than 0.05.
Exam move
Memorise both statements word for word, hypotheses included, and practise the tilting proof until you can write it unprompted in five minutes. Then work three problem types in rotation: a uniqueness of root question using Rolle with the Intermediate Value Theorem, an inequality from a derivative bound, and a counterexample showing a hypothesis cannot be dropped.
Mark each solution by checking that the hypotheses were verified before the theorem was quoted.
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