Source: MTH 132 Chapter 3 Lecture Notes, Michigan State University
Difficulty: Intermediate | Prerequisites: Continuity, differentiability, Chapter 3 Section 1 (critical numbers, EVT).
The Mean Value Theorem (MVT) is one of the most important theoretical results in all of calculus. It connects the average rate of change of a function over an interval to an instantaneous rate of change at some point inside that interval. Rolle's Theorem is a special case that sets the stage. These theorems are "existence" results: they guarantee a particular point exists but do not hand you its value directly. You will use the MVT to prove why certain equations have limited solutions and to justify corollaries about constant functions.
If a function is continuous on [a, b] and differentiable on (a, b), there is at least one point c in between where the tangent line's slope equals the slope of the secant line through the endpoints. Rolle's Theorem is the special case where the endpoints have the same function value, so the tangent line at c is horizontal.
Rolle's Theorem
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists at least one c in (a, b) with f'(c) = 0. Think of it as: if the function starts and ends at the same height, it must have a horizontal tangent somewhere in between.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), then there exists a c in (a, b) such that f'(c) = [f(b) - f(a)] / (b - a). In simple terms, somewhere between a and b the instantaneous slope equals the average slope.
Secant line
The straight line connecting the points (a, f(a)) and (b, f(b)). Its slope is [f(b) - f(a)] / (b - a).
Tangent line
The line that touches the curve at a single point and has slope equal to the derivative at that point.
Corollary: zero derivative implies constant
If f'(x) = 0 for all x in (a, b), then f(x) is constant on (a, b).
Corollary: equal derivatives implies constant difference
If f'(x) = g'(x) for all x in (a, b), then f(x) = g(x) + c for some constant c.
Three hypotheses must all hold:
f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).
f(a) = f(b).
Conclusion: there exists at least one c in (a, b) where f'(c) = 0.
Geometrically, if the curve starts and ends at the same height, it must turn around somewhere, and at that turning point the tangent line is horizontal. Rolle's Theorem is a special case of the MVT where the secant line is flat.
Two hypotheses:
f is continuous on [a, b].
f is differentiable on (a, b).
Conclusion: there exists at least one c in (a, b) such that
f'(c) = [f(b) - f(a)] / (b - a)
The left side is the slope of the tangent at c. The right side is the slope of the secant line joining (a, f(a)) to (b, f(b)). The theorem says there is a place in the interval where the tangent line is parallel to the secant line.
Both Rolle's Theorem and the MVT are existence theorems. They tell you "there exists" a number c with certain properties, but they do not tell you what c is or how to find it. To actually find c, you set up the equation f'(c) = [f(b) - f(a)] / (b - a) and solve.
Corollary 1: If f'(x) = 0 for all x in an interval (a, b), then f(x) is constant on (a, b).
This makes intuitive sense: if the slope is always zero, the function is flat, so the output never changes.
Corollary 2: If f'(x) = g'(x) for all x in (a, b), then f(x) = g(x) + c for some constant c.
In other words, two functions with the same derivative everywhere on an interval can only differ by a constant. This is why indefinite integrals always include "+ C."
Example: f(x) = x^2 satisfies f'(x) = 2x. Another function with the same derivative is f(x) = x^2 + 47. Any function whose derivative is 2x has the form x^2 + C.
Example 1: f(x) = x^2 + 3x + 5 on [0, 1]
f is a polynomial, so it is continuous and differentiable everywhere. The MVT applies.
f'(x) = 2x + 3. f(1) = 9, f(0) = 5.
[f(1) - f(0)] / (1 - 0) = 4.
Set f'(c) = 4: 2c + 3 = 4, so c = 1/2. Since 1/2 is in (0, 1), this is the answer.
Example 2: f(x) = 1/x on [1, 3]
f is continuous and differentiable on [1, 3] (no discontinuity in this interval). The MVT applies.
f'(x) = -1/x^2. f(3) = 1/3, f(1) = 1.
[f(3) - f(1)] / (3 - 1) = (1/3 - 1)/2 = -1/3.
Set -1/c^2 = -1/3: c^2 = 3, so c = sqrt(3). Since sqrt(3) is approximately 1.73 and lies in (1, 3), this works.
Example 3: f(x) = |x| on [-2, 2]
f is continuous on [-2, 2], but f'(0) does not exist (corner at x = 0). Since 0 is in (-2, 2), f is not differentiable on the open interval. The MVT does not apply.
Example 4: f(x) = cube root of x on [0, 1]
f(x) = x^(1/3) is continuous on [0, 1]. f'(x) = (1/3)x^(-2/3) = 1/(3 * x^(2/3)). This is undefined at x = 0, but 0 is an endpoint, not in (0, 1). So f is differentiable on (0, 1), and the MVT applies.
[f(1) - f(0)] / (1 - 0) = 1. Set (1/3)c^(-2/3) = 1, so c^(-2/3) = 3, c^(2/3) = 1/3, c = (1/3)^(3/2) = 1/sqrt(27).
Example 5: Proof using the MVT
Show that 3x + 2cos(x) + 5 = 0 has at most one solution.
Let f(x) = 3x + 2cos(x) + 5. Suppose f has two roots c1 and c2. Then f(c1) = f(c2) = 0.
By Rolle's Theorem, there exists a d between c1 and c2 where f'(d) = 0.
f'(x) = 3 - 2sin(x). Since -1 <= sin(x) <= 1, we have f'(x) = 3 - 2sin(x) >= 3 - 2 = 1 > 0 for all x.
So f'(d) >= 1, which contradicts f'(d) = 0. Therefore, f cannot have two roots.
Example 6: f(x) = 2sin(x)cos(x) on [0, pi]
sin(x) and cos(x) are continuous and differentiable everywhere, so their product is too. The MVT applies.
f(pi) = 2sin(pi)cos(pi) = 0. f(0) = 0. So f(a) = f(b), and Rolle's Theorem applies as a special case.
f'(x) = 2(cos^2(x) - sin^2(x)) = 2cos(2x). Set 2cos(2c) = 0, so cos(2c) = 0, giving 2c = pi/2 or 2c = 3pi/2, so c = pi/4 or c = 3pi/4.
Students often try to apply the MVT without checking both hypotheses. Always verify continuity on [a, b] and differentiability on (a, b) first.
Students confuse where the hypotheses are checked. Continuity is needed on the closed interval [a, b]; differentiability is needed on the open interval (a, b). Endpoints do not need to be differentiable.
Students sometimes think the MVT tells you the exact value of c. It guarantees c exists but does not hand it to you. You find c by solving the equation.
Students mix up Rolle's Theorem with the MVT. Rolle's is the special case where f(a) = f(b), which gives f'(c) = 0. The full MVT gives f'(c) = [f(b) - f(a)]/(b - a).
Expect "verify the hypotheses and find c" problems: you check continuity, check differentiability, compute the secant slope, set f'(c) equal to it, and solve.
Expect "does the MVT apply?" problems where a function fails one hypothesis (e.g., |x| on [-2, 2] is not differentiable at x = 0).
Proof by contradiction using Rolle's Theorem (showing a function has at most one root) appears on exams.
The corollaries (f' = 0 implies f constant, f' = g' implies f = g + C) are tested conceptually and connect directly to the idea of antiderivatives and the "+C" in integration.
True or False: The MVT requires f to be differentiable on the closed interval [a, b].
Answer: False. Differentiability is required on the open interval (a, b).
Fill in the blank: Rolle's Theorem adds the extra condition that f(a) = ____, and its conclusion is that f'(c) = ____.
Answer: f(b); 0.
True or False: If f'(x) = 0 for all x in (a, b), then f is constant on (a, b).
Answer: True.
Fill in the blank: The MVT says the slope of the ____ line at some c equals the slope of the ____ line through the endpoints.
Answer: tangent; secant.
Q: Can the MVT be applied to f(x) = |x| on [-2, 2]? Why or why not?
A: No. f(x) = |x| is not differentiable at x = 0, and 0 is in the open interval (-2, 2). The differentiability hypothesis fails.
Q: Consider f(x) = x + 11/x on [1/4, 4]. Verify the MVT applies and find all c values satisfying its conclusion.
A: f is discontinuous at x = 0, but 0 is not in [1/4, 4], so f is continuous there. f'(x) = 1 - 11/x^2, which exists on (1/4, 4). MVT applies. The secant slope is [f(4) - f(1/4)] / (4 - 1/4) = [4 + 11/4 - (1/4 + 44)] / (15/4) = -10. Setting 1 - 11/c^2 = -10 gives c^2 = 1, so c = 1 (the value in (1/4, 4)).
Q: Consider g(x) = sqrt(7 - x) on [-7, 7]. Verify the MVT hypotheses and find all c.
A: g is continuous on [-7, 7] (7 - x >= 0 throughout). g'(x) = -1/(2sqrt(7 - x)), which is undefined at x = 7, but that is an endpoint. g is differentiable on (-7, 7). MVT applies. Secant slope = [g(7) - g(-7)]/(7 - (-7)) = [0 - sqrt(14)]/14 = -sqrt(14)/14. Setting -1/(2sqrt(7 - c)) = -sqrt(14)/14, solving gives 7 - c = 49/14 = 7/2, so c = 7/2.
Q: Show that 3x + 2cos(x) + 5 = 0 has at most one solution.
A: Let f(x) = 3x + 2cos(x) + 5. Suppose f has two roots c1, c2. By Rolle's Theorem, f'(d) = 0 for some d between them. But f'(x) = 3 - 2sin(x) >= 1 for all x, a contradiction. So at most one root exists.
The MVT is the theoretical backbone of the First Derivative Test and the Increasing/Decreasing Test (Chapter 3, Section 3). Those results follow from the MVT.
The corollary about equal derivatives connects directly to antiderivatives and indefinite integrals (Chapter 4), explaining why the general antiderivative of a function includes "+ C."
Rolle's Theorem is used in proofs throughout later calculus courses, including in proving the uniqueness of solutions to differential equations.
Mean Value Theorem, MVT, Rolle's Theorem, secant line, tangent line parallel to secant, average rate of change, instantaneous rate of change, existence theorem, zero derivative constant function, equal derivatives differ by constant, proof by contradiction, applications of differentiation, MTH 132 Chapter 3, Calculus I MSU