Rolle's theorem and the Mean Value Theorem, the proof by subtracting the chord, the two hypotheses and counterexamples showing both are needed, and the three corollaries — a zero derivative gives a constant, equal derivatives differ by a constant, and the derivative's sign determines increase or decrease.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
The Mean Value Theorem
Objectives
Five outcomes. The last is what Section 4.5 assumes throughout, and it cannot be proved any other way.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 329-338 — the section these objectives are drawn from
Warm-up
Section 3.2 read the sign of a derivative off a graph and concluded the function was rising or falling. That reading has been used ever since without proof.
Discussion prompt
Why should a positive derivative at every point of an interval force the function to be larger at the right end than at the left? The derivative only describes single instants.
Hint: You need a statement connecting instantaneous behaviour to behaviour over a whole interval.
Answer:
Nothing proved so far bridges the gap. A derivative describes what happens AT a point, and the claim is about what happens ACROSS an interval — and no result yet connects the two.
\[ f' > 0 \text{ everywhere on } (a,b) \;\Longrightarrow\;? \; f(b) > f(a) \]
The Mean Value Theorem is exactly that bridge. It says the average rate across an interval is attained instantaneously somewhere inside, which converts a statement about every instant into one about the endpoints — and the whole of Section 4.5 depends on it.
Concept
For a function continuous on a closed interval and differentiable inside it, there is some interior point where the instantaneous rate equals the average rate over the whole interval. Geometrically, some tangent is parallel to the chord.
the Mean Value Theorem — If f is continuous on the closed interval from a to b and differentiable on the open interval, then there is a c strictly between them with f prime of c equal to the difference in values divided by the difference in inputs.
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \quad \text{for some } c \in (a,b) \]
The theorem is an existence statement like those of Sections 2.4 and 4.3: it promises the point c is there and says nothing about where. Its value is entirely in what can be deduced from knowing such a point exists.
Figure (svg): A curve with the chord between its endpoints and a parallel tangent somewhere between
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 329-332
Section
Section 1
Concept
If a function is continuous on a closed interval, differentiable inside, and takes the same value at both ends, then its derivative vanishes somewhere strictly between. A curve that returns to its starting height must turn around.
Rolle's theorem — If f is continuous on a closed interval, differentiable on its interior, and f of a equals f of b, then there is an interior c with f prime of c equal to zero.
\[ f(a)=f(b) \;\Longrightarrow\; \exists c \in (a,b): f'(c)=0 \]
The proof combines two earlier results. The Extreme Value Theorem guarantees a maximum and a minimum exist, and Fermat's theorem forces a zero derivative at whichever of them is interior — and at least one must be, unless the function is constant.
Figure (svg): A curve with equal values at both endpoints and a horizontal tangent somewhere between
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 329-331 — Rolle's theorem
Picture it
A parabola with equal values at two inputs.
Figure (svg): A curve with equal values at both endpoints and a horizontal tangent somewhere between
The curve dips and returns, and at the bottom of the dip the tangent is level. That is the theorem, and the picture is essentially the proof.
Worked example
Example 4.19. Check the hypotheses, then find c.
\[ \text{Verify Rolle's theorem for } f(x)=x^{2}-4x+3 \text{ on } [1,3] \text{ and find } c. \]
Check continuity and differentiability
Why: A polynomial.
Check the endpoint values agree
Why: Evaluate both.
\[ f(1) = 0\text{ and } f(3) = 0 \]
Conclude the theorem applies
Why: All three hypotheses hold.
\[ \text{some } c\text{ has } f'(c) = 0 \]
Differentiate and solve
Why: Set to zero.
\[ 2 x - 4 = 0,\text{ so } x = 2 \]
Check c is strictly inside
Why: Two lies between 1 and 3.
\[ c = 2 \]
Figure (svg): The solution to Worked example applying Rolle's theorem shown as a ladder of expressions, one row per legal move
\[ c = 2, \quad f'(2) = 0 \]
Verify: check the value at c and the picture
Why: The value there is 4 minus 8 plus 3, which is negative 1 — the parabola's vertex, and its lowest point on the interval. That the guaranteed c turns out to be the minimum is no accident: the proof locates c precisely as an interior extremum. Note that the theorem said such a c exists without telling us where, and finding it required solving the derivative equation separately.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 330-331
Sorting
Check continuity, differentiability inside, and equal endpoints.
Sort into buckets
Sort each case.
The squaring function on the interval from 0 to 2 fails only because its endpoint values differ, 0 against 4 — a reminder that the equal-endpoints condition is a genuine hypothesis and not a formality.
Worked example
Checkpoint 4.19. The proof in four lines.
\[ \text{Prove Rolle's theorem from earlier results.} \]
Apply the Extreme Value Theorem
Why: Continuity on a closed interval.
Consider the case where both are at endpoints
Why: Then both equal the same value.
Otherwise one extremum is interior
Why: It occurs strictly inside.
Apply Fermat's theorem there
Why: f is differentiable inside.
\[ f'(c) = 0 \]
Conclude
Why: Both cases give a c.
Figure (svg): The solution to Worked example why the theorem is true shown as a ladder of expressions, one row per legal move
\[ \text{EVT} + \text{Fermat} \;\Longrightarrow\; \text{Rolle} \]
Verify: check the constant case is genuinely needed
Why: If both extrema sit at the endpoints then the maximum and minimum are equal, so the function takes that one value throughout and is constant — in which case the derivative vanishes everywhere and any interior c works. Without treating that case the argument would have a gap, since Fermat's theorem needs an INTERIOR extremum and there might be none. The two-case structure is what makes the proof complete.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 331-331
Trap
\[ f(x)=|x| \text{ on } [-1,1]: \; f(-1)=f(1)=1 \]
Conclude a horizontal tangent exists inside
Why: The student checks only the equal endpoints.
\[ \text{but } f'(x) = \pm 1 \text{ everywhere it exists} \]
The derivative is never zero. The function is not differentiable at the corner, so the theorem does not apply.
\[ \text{check: continuous on } [a,b], \text{ differentiable on } (a,b), \; f(a)=f(b) \]
Verify all three hypotheses before invoking the conclusion
Why: The absolute value fails the middle one at exactly one point, and that is enough.
A single point of non-differentiability defeats the theorem entirely, which is why the hypothesis is stated for the whole open interval. The corner is where the turning happens, and it is precisely the place a derivative would have had to vanish.
Fill the middle
The parabola from the worked example, with its derivative set to zero.
Fill in the blanks
2x - 4 = 0 \;\Longrightarrow\; c = 2
Why: The derivative vanishes at 2, which lies strictly inside the interval as the theorem requires. It is the parabola's vertex, and the proof locates c precisely as an interior extremum.
Ranking
Rolle's theorem from earlier results.
Put in order
Why: Step b is the case that is easy to overlook and genuinely needed, since Fermat's theorem requires an interior extremum and a constant function might have none. Every earlier theorem in Chapter 4 feeds into this one.
Prediction
Commit before reasoning.
Predict first
What goes wrong if f(a) does not equal f(b)?
Correct: The curve need not turn around, so no level tangent is forced.
\[ x^{2} \text{ on } [0,2]: \; f' = 2x > 0 \text{ on } (0,2) \]
Why: The squaring function on the interval from 0 to 2 climbs steadily from 0 to 4 with a positive derivative throughout, and its derivative is zero only at the left endpoint rather than strictly inside. Nothing forces a turn when the ends are at different heights. The Mean Value Theorem is exactly the generalisation that handles this case, replacing 'horizontal' with 'parallel to the chord'.
Section
Section 2
Concept
Dropping the requirement that the endpoints agree gives the general theorem. The chord's slope is the average rate of change, and some interior tangent has exactly that slope.
average and instantaneous rate — The average rate of change over an interval is the difference in values divided by the difference in inputs — the chord's slope. The theorem says some instantaneous rate inside equals it.
\[ \frac{f(b)-f(a)}{b-a} = f'(c) \]
Read as a statement about motion it is striking: an object's average velocity over a journey is its actual velocity at some instant of that journey, whatever else it did.
Figure (svg): A curve with the chord between its endpoints and a parallel tangent somewhere between
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 331-334 — the Mean Value Theorem
Picture it
A parabola with the chord between its endpoints.
Figure (svg): A curve with the chord between its endpoints and a parallel tangent somewhere between
The chord's slope is the average rate over the whole interval, and at the marked point the tangent has exactly that slope. Sliding the chord parallel until it touches locates c.
Worked example
Example 4.21. Compute the chord's slope, then solve.
\[ \text{Find all } c \text{ guaranteed by the theorem for } f(x)=x^{2} \text{ on } [0,4]. \]
Check the hypotheses
Why: A polynomial.
Compute the chord's slope
Why: The average rate.
\[ \frac{16 - 0}{4 - 0} = 4 \]
Set the derivative equal to it
Why: The theorem's equation.
\[ 2 c = 4 \]
Solve
Why: Divide.
\[ c = 2 \]
Check c is strictly inside
Why: Two lies between 0 and 4.
Figure (svg): The solution to Worked example finding the guaranteed point shown as a ladder of expressions, one row per legal move
\[ c = 2, \quad f'(2) = 4 = \frac{f(4)-f(0)}{4-0} \]
Verify: check whether the midpoint is a coincidence
Why: For any parabola the guaranteed point is always the midpoint of the interval, because the derivative is linear and the average of a linear function over an interval is its value at the centre. For other functions c is generally not the midpoint — for the cubing function on the interval from 0 to 3 it is the root of 3, about 1.73. So the midpoint here reflects the parabola's symmetry rather than anything general.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 332-333
Fill the middle
The squaring function on the interval from zero to four.
Fill in the blanks
\frac4___ = \frac______ = ___
Why: The average rate is 4, and setting the derivative 2c equal to it gives c equal to 2. The chord's slope is always the first thing to compute.
Worked example
Example 4.22. Two measurements force a conclusion about an instant.
\[ \text{A car passes two cameras } 39 \text{ miles apart in } 36 \text{ minutes. The limit is } 60 \text{ mph. Was it speeding?} \]
Convert to consistent units
Why: Thirty-six minutes is 0.6 hours.
\[ t\text{ from } 0\text{ to } 0.6 \]
Compute the average speed
Why: Distance over time.
\[ \frac{39}{0.6} = 65 \text{mph} \]
Check the hypotheses
Why: Position is continuous and differentiable in time.
Apply it
Why: Some instant has that speed.
\[ s'(c) = 65\text{ for some } c \]
Compare with the limit
Why: Sixty-five exceeds 60.
Figure (svg): The speeding argument: an average speed above the limit forces an instant above it
\[ s'(c) = 65 > 60 \text{ for some } c \]
Verify: notice what was and was not observed
Why: No one measured an instantaneous speed at all — only two positions and two times. The theorem converts that into a certain statement about an instant, which is why average-speed enforcement is legally sound. Note also what it does NOT give: it says nothing about WHEN the car was doing 65, so a ticket can assert the offence occurred without locating it. That is precisely the shape of an existence theorem.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 333-334
Error analysis
A student applies the theorem to a journey.
Annotate
On: \( \text{average } 65 \text{ mph over } 36 \text{ min} \;\Longrightarrow\; \text{the car did } 65 \text{ at the halfway point} \)
Like the Intermediate Value Theorem and the Extreme Value Theorem, this is an existence result. It promises the instant is there and locates it only when the derivative equation is solved explicitly.
Matching
The same theorem in three languages.
Match the pairs
Why: All three readings are the same theorem, and each is useful in a different setting. The last row is the honest caveat that applies to all of them: existence without location, exactly like the Intermediate Value Theorem.
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Like every existence theorem in this course, it promises c exists and is silent about its location. Finding c requires solving the equation f prime of c equals the chord's slope, which is a separate computation the theorem neither performs nor guarantees is tractable.
Prediction
Commit before reasoning.
Predict first
Two cameras record positions and times. How can that prove an instantaneous speed was exceeded?
Correct: The theorem converts the measured average into a guaranteed instantaneous value.
\[ \frac{39}{0.6} = 65 = s'(c) \text{ for some instant } c \]
Why: Position is a continuous, differentiable function of time, so the theorem applies and the average of 65 mph must have been attained at some instant. That is why average-speed enforcement is legally defensible from two measurements alone. Average and instantaneous speed are certainly not generally equal — the whole content is that they coincide at least once, which is a far weaker and far more useful claim.
Section
Section 3
Concept
Define a new function as the original minus the chord. It agrees with the chord at both endpoints, so the new function takes the value zero at each — and Rolle's theorem applies to it.
the auxiliary function — The difference between f and the linear function whose graph is the chord. It vanishes at both endpoints, so Rolle's theorem gives an interior point where its derivative is zero, which translates back into the Mean Value Theorem.
\[ g(x) = f(x) - \left[f(a) + \frac{f(b)-f(a)}{b-a}(x-a)\right] \]
The move is exactly a change of viewpoint: tilt the picture until the chord is level, apply the level-chord theorem, and tilt back. Nothing is lost because subtracting a linear function subtracts a constant from every derivative.
Figure (svg): The proof idea: subtract the chord to reduce the Mean Value Theorem to Rolle's
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 332-334 — the proof of the Mean Value Theorem
Picture it
The original and the difference with the chord.
Figure (svg): The proof idea: subtract the chord to reduce the Mean Value Theorem to Rolle's
On the right the endpoints sit at the same height, so Rolle's theorem gives a horizontal tangent — and translating back, that horizontal tangent is a tangent parallel to the original chord.
Worked example
Example 4.23. Four lines from Rolle to the general theorem.
\[ \text{Prove the Mean Value Theorem from Rolle's theorem.} \]
Define the auxiliary function
Why: f minus the chord.
\[ g(x) = f(x) - f(a) - m(x - a),\text{ with } m\text{ the chord slope} \]
Check its endpoint values
Why: Both come out zero.
\[ g(a) = 0\text{ and } g(b) = 0 \]
Check the hypotheses transfer
Why: A linear function is continuous and differentiable.
Apply Rolle's theorem to g
Why: All three conditions hold.
\[ g'(c) = 0\text{ for some interior } c \]
Translate back
Why: g' is f' minus the constant m.
\[ f'(c) = m \]
Figure (svg): The proof idea: subtract the chord to reduce the Mean Value Theorem to Rolle's
\[ g'(c) = f'(c) - m = 0 \;\Longrightarrow\; f'(c) = m \]
Verify: check that g really vanishes at both ends
Why: At x equal to a the bracket gives f of a plus zero, so g of a is f of a minus f of a, which is 0. At x equal to b the bracket gives f of a plus the chord slope times b minus a, which is exactly f of b — so g of b is 0 as well. Both endpoint values vanish, which is precisely what Rolle's theorem needs. The subtraction was designed to make that happen, and nothing else about it matters.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 333-334
Fill the middle
The auxiliary function evaluated at the left endpoint.
Fill in the blanks
g(a) = f(a) - f(a) - m(a-a) = 0
Why: Both terms cancel and the last is zero, so g vanishes at the left endpoint. The same computation at b gives zero as well, which is exactly the condition Rolle's theorem needs.
Worked example
Checkpoint 4.23. The derivative shifts by a constant.
\[ \text{Explain why } g'(x) = f'(x) - m \text{ and why that is what the proof needs.} \]
Differentiate the auxiliary function
Why: The bracket is linear in x.
\[ g'(x) = f'(x) - m \]
Note the constant term vanishes
Why: f(a) is a constant.
Set the derivative to zero
Why: Rolle's conclusion.
\[ f'(c) - m = 0 \]
Solve
Why: The theorem's statement.
\[ f'(c) = m \]
Figure (svg): The solution to Worked example why subtracting a line is safe shown as a ladder of expressions, one row per legal move
\[ g' = f' - m \;\Longrightarrow\; g'(c)=0 \iff f'(c)=m \]
Verify: confirm the translation is exact rather than approximate
Why: The relation between g prime and f prime holds at every input, not just at c, so a zero of one corresponds exactly to a point where the other equals m. Nothing is lost or approximated in the translation. That exactness is why the tilting device works: subtracting a linear function shifts every slope by the same amount, so the geometry is genuinely rotated rather than distorted.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 334-334
Trap
\[ g(x) = f(x) - mx \]
Subtract only the slope term
Why: The student omits the constant that makes the line pass through the first endpoint.
\[ g(a) = f(a) - ma \ne 0 \text{ in general} \]
Rolle's theorem needs the two endpoint values to AGREE, and this g does not deliver that.
\[ g(x) = f(x) - f(a) - m(x-a) \;\Longrightarrow\; g(a) = g(b) = 0 \]
Subtract the whole chord, anchored at the first endpoint
Why: It is the line through both endpoints, not merely a line of the right slope.
Any line of the correct slope would give the same derivative relation, and Rolle's theorem would still apply provided the two endpoint values agreed — which they do for any such line, since both are shifted equally. The anchored form simply makes the common value zero, which is tidiest.
Ranking
The Mean Value Theorem from Rolle's.
Put in order
Why: Step c is the whole point of the construction and worth verifying explicitly the first time. Step e is exact rather than approximate, because subtracting a linear function shifts every derivative by the same constant.
Two truths and a lie
All three are about the proof.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The proof uses only continuity on the closed interval and differentiability inside, exactly as the theorem states. It applies to any such function, including ones with no formula at all — which matters, since the speeding argument applies to a car's position function that nobody has written down.
Prediction
Commit before reasoning.
Predict first
Why does subtracting the chord not change which points have tangents parallel to it?
Correct: Because subtracting a line reduces every slope by the same constant.
\[ f'(x) = m \iff g'(x) = 0, \text{ at every } x \]
Why: Every tangent's slope drops by exactly m, so a tangent that was parallel to the chord now has slope zero and one that was not still does not. The correspondence is exact and applies at every input, which is why the translation in the final step is an equivalence rather than an approximation. The function certainly changes; what is preserved is the relationship between slopes, and that is all the proof needs.
Section
Section 4
Concept
Both conditions are needed, and they are asked for on different sets. Continuity is required on the closed interval including the endpoints; differentiability only on the open interior.
the asymmetric hypotheses — Continuity is demanded on the closed interval so the endpoint values are meaningful; differentiability only inside, because the conclusion concerns an interior point and endpoint derivatives are never needed.
\[ f \text{ continuous on } [a,b], \text{ differentiable on } (a,b) \]
The asymmetry is deliberate and useful. The square root on the interval from 0 to 4 has no derivative at its left endpoint and the theorem still applies, because differentiability is only required strictly inside.
Figure (svg): The two hypotheses failing, each producing a counterexample
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 331-335 — hypotheses of the theorem
Picture it
A corner and a jump.
Figure (svg): The two hypotheses failing, each producing a counterexample
The absolute value has a level chord and no level tangent; the step function has a sloping chord and every tangent flat. Each counterexample isolates exactly one hypothesis.
Worked example
Example 4.24. A corner defeats the theorem.
\[ \text{Show the theorem fails for } f(x)=|x| \text{ on } [-1,1]. \]
Compute the chord's slope
Why: Equal endpoint values.
\[ \frac{1 - 1}{2} = 0 \]
Look for a point with that derivative
Why: The theorem would need f'(c) = 0.
Examine the derivative
Why: It is 1 or negative 1 wherever it exists.
Diagnose
Why: The derivative fails at the corner.
\[ \text{not differentiable at } 0 \]
Conclude
Why: The hypothesis fails, and so does the conclusion.
Figure (svg): The solution to Worked example differentiability failing shown as a ladder of expressions, one row per legal move
\[ f'(x) = \pm 1 \text{ everywhere it exists} \]
Verify: confirm the failure is at exactly one point
Why: The absolute value is differentiable at every input except the origin, so the hypothesis fails at a single point out of infinitely many — and that is enough to destroy the conclusion entirely. The theorem's hypotheses are not approximate conditions to be roughly satisfied; a single exception defeats them. Note also that the corner is precisely where a horizontal tangent would have had to be.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 334-335
Sorting
Continuity on the closed interval, differentiability inside.
Sort into buckets
Sort each case.
The cube root fails because its vertical tangent is at the origin, which is INSIDE the interval — unlike the square root, whose failure is at an endpoint where nothing is required. The location of the failure decides the case.
Worked example
Checkpoint 4.24. Differentiability is only needed inside.
\[ \text{Does the theorem apply to } f(x)=\sqrt{x} \text{ on } [0,4]? \]
Check continuity on the closed interval
Why: The root is continuous on its domain.
Check differentiability at the left endpoint
Why: There is a vertical tangent.
\[ f'(0)\text{ does not exist} \]
Recall what is actually required
Why: Only the open interval.
\[ \text{differentiable on } (0, 4) \]
Check that
Why: The derivative exists for every positive x.
Conclude
Why: Both hypotheses are satisfied.
Figure (svg): The solution to Worked example the asymmetry put to use shown as a ladder of expressions, one row per legal move
\[ \text{continuous on } [0,4], \text{ differentiable on } (0,4) \]
Verify: find the guaranteed c
Why: The chord's slope is 2 over 4, or one half, and setting one over twice the root of c equal to one half gives c equal to 1 — comfortably inside the interval. Had differentiability been demanded on the closed interval, this perfectly ordinary function would have been excluded for no good reason. The asymmetry in the hypotheses is what keeps the theorem widely applicable.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 335-335
Error analysis
A student rejects an application of the theorem.
Annotate
On: \( \sqrt{x} \text{ on } [0,4]: \; f'(0) \text{ does not exist, so the theorem fails} \)
The two hypotheses are asked for on different sets, and the difference is not cosmetic. Demanding differentiability at the endpoints would exclude the square root, the cube root and many other perfectly ordinary functions.
Fill the middle
The theorem's hypotheses, asked for on different intervals.
Fill in the blanks
\text(a,b) [a,b], \quad \text___ ___
Why: Differentiability is only required strictly inside, which is what allows the square root on the interval from 0 to 4 despite its vertical tangent at the left end.
Two truths and a lie
All three are about the hypotheses.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The absolute value is continuous everywhere on the interval from negative 1 to 1 and the conclusion still fails, because it is not differentiable at the origin. Both hypotheses are needed, and the two counterexamples in this section drop one each to prove it.
Prediction
Commit before reasoning.
Predict first
Why is differentiability required only on the open interval?
Correct: Because the conclusion is about an interior point, so endpoint derivatives play no part.
\[ c \in (a,b): \text{ the conclusion never mentions } a \text{ or } b \text{ as derivative points} \]
Why: The theorem produces a c strictly between a and b, and the proof only differentiates inside. Demanding more would exclude functions like the square root for no benefit, since their endpoint behaviour is irrelevant to the argument. Endpoint derivatives can perfectly well exist — for a polynomial they always do — and the point is simply that they are not needed.
Section
Section 5
Concept
Three consequences follow immediately. A zero derivative throughout an interval makes the function constant; two functions with equal derivatives differ by a constant; and the sign of the derivative determines whether the function increases or decreases.
the corollaries — A function with zero derivative on an interval is constant there. Two functions with the same derivative differ by a constant. A positive derivative gives an increasing function and a negative one a decreasing function.
\[ f' > 0 \text{ on } (a,b) \;\Longrightarrow\; f \text{ increasing on } (a,b) \]
The second corollary is what makes antiderivatives well behaved in Chapter 5. It says the constant of integration is the only ambiguity, and that fact has to be proved rather than assumed.
Figure (svg): The three corollaries, each stating what a derivative condition forces on the function
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 335-338 — corollaries of the Mean Value Theorem
Picture it
What each derivative condition forces.
Figure (svg): The three corollaries, each stating what a derivative condition forces on the function
Every one of these is used routinely and looks self-evident, and none can be proved without the theorem. That is what makes this section foundational rather than decorative.
Worked example
Example 4.25. The theorem does the whole job.
\[ \text{Prove that if } f' > 0 \text{ on an interval then } f \text{ is increasing there.} \]
Take any two inputs in the interval
Why: With the first smaller.
\[ x 1 < x 2 \]
Apply the theorem on the interval between them
Why: The hypotheses hold there.
\[ f(x 2) - f(x 1) = f'(c) (x 2 - x 1) \]
Examine the signs of the two factors
Why: Both are positive.
\[ f'(c) > 0\text{ and } x 2 - x 1 > 0 \]
Conclude about the product
Why: Positive times positive.
\[ f(x 2) - f(x 1) > 0 \]
State
Why: The larger input gives the larger output.
Figure (svg): The argument that a positive derivative makes a function increasing
\[ f(x_{2}) > f(x_{1}) \text{ whenever } x_{1} < x_{2} \]
Verify: notice what the argument needed
Why: The derivative's sign was known at every point of the interval, but the argument used it at only ONE point — the c the theorem supplied. That is the leverage: a hypothesis about every instant is converted into a conclusion about two endpoints by way of a single unspecified intermediate point. Without the theorem there is no route from 'the slope is positive everywhere' to 'the value is larger at the right end'.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 336-337
Matching
What each derivative fact forces.
Match the pairs
Why: Every row needs the Mean Value Theorem, and the second is the one Chapter 5 leans on: it says two antiderivatives of the same function can only differ by a constant, which is exactly what the constant of integration records.
Worked example
Checkpoint 4.25. The corollary Chapter 5 depends on.
\[ \text{Prove that if } f'=g' \text{ on an interval then } f-g \text{ is constant.} \]
Define the difference
Why: A new function.
\[ h = f - g \]
Differentiate it
Why: The derivatives cancel.
\[ h' = f' - g' = 0 \]
Apply the first corollary
Why: A zero derivative on an interval.
Translate back
Why: The difference is that constant.
\[ f - g = C \]
State
Why: They differ by a constant.
\[ f = g + C \]
Figure (svg): The solution to Worked example equal derivatives differ by a constant shown as a ladder of expressions, one row per legal move
\[ f' = g' \;\Longrightarrow\; f = g + C \]
Verify: check that the interval hypothesis is essential
Why: On a domain in two separate pieces this fails: the function equal to 1 on the positive numbers and 2 on the negative ones has zero derivative throughout its domain and is not constant. The corollary needs a single interval, because the theorem needs two points to be joinable by an interval inside the domain. This is why the constant of integration in Chapter 5 is a single constant only on a connected interval.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 337-338
Trap
\[ f(x) = \frac{|x|}{x}: \; f'(x) = 0 \text{ wherever } f \text{ is defined} \]
Conclude the function is constant
Why: The student applies the first corollary.
\[ \text{but } f = -1 \text{ on the left and } +1 \text{ on the right} \]
The domain is two separate intervals, and the function takes a different constant value on each.
\[ f' = 0 \text{ on a single INTERVAL } \;\Longrightarrow\; f \text{ constant there} \]
Check the domain is one connected interval
Why: The theorem joins two points by an interval, which is impossible across a gap.
The consequence appears again in Chapter 5: the antiderivative of one over x is the logarithm of the absolute value PLUS A CONSTANT, but the constant may differ on the two sides of the origin, because the domain is disconnected.
Fill the middle
The theorem applied between two inputs, with both factors positive.
Fill in the blanks
f(x_>) - f(x____) = f'(c)(x____-x____) ___ 0
Why: Both factors are positive, so the product is positive and the value at the larger input exceeds the value at the smaller. That is the definition of increasing.
Two truths and a lie
All three are about the corollaries.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Each corollary connects information at every point to a conclusion about two points, and there is no route between them without the Mean Value Theorem. That the statements feel obvious is exactly why the theorem is easy to undervalue — the picture is compelling long before the proof exists.
Prediction
Commit before reasoning.
Predict first
Which corollary guarantees that the constant of integration is the only ambiguity in an antiderivative?
Correct: That two functions with equal derivatives differ by a constant.
\[ F' = G' = f \;\Longrightarrow\; F = G + C: \text{ the constant is the ONLY freedom} \]
Why: An antiderivative of a given function is one whose derivative is that function, and there are many of them. The corollary says any two of them have the same derivative and therefore differ by a constant — so the family of all antiderivatives is exactly one of them plus an arbitrary constant. Without this, writing plus C would be a hopeful convention rather than a complete description, and the whole notation of Chapter 5 would be unjustified.
Comparison
Fill the blanks. Each guarantees something exists and locates nothing.
Comparison matrix
| Theorem | What it guarantees | Section |
|---|---|---|
| Intermediate Value | a value between two others is attained | 2.4 |
| Extreme Value | a maximum and a minimum are attained | 4.3 |
| Rolle | a horizontal tangent, given equal endpoints | 4.4 |
| Mean Value | a tangent parallel to the chord | 4.4 |
The chain is cumulative: the Extreme Value Theorem and Fermat's theorem prove Rolle's, and Rolle's proves the Mean Value Theorem, which in turn proves everything Section 4.5 will assert.
Pattern
Given a function and an interval, to apply the theorem.
Steps one and two are asked on different sets and that difference matters — a function may fail to be differentiable at an endpoint and still satisfy the hypotheses completely.
Stewart, Calculus: Early Transcendentals 8e, §4.2 The Mean Value Theorem §4.2, pp. 287-292
Check
Rolle's theorem. Find the guaranteed point.
Check your understanding
For f(x) = x^2 - 4x + 3 on [1,3], what value does Rolle's theorem guarantee?
Answer: A
Why: f(1) = f(3) = 0, and 2x - 4 vanishes at x = 2, which is strictly inside.
Check
The Mean Value Theorem. Chord slope first.
Check your understanding
For f(x) = x^2 on [0,4], what c does the theorem guarantee?
Answer: A
Why: The chord's slope is 16/4 = 4, and 2c = 4 gives c = 2.
Check
The corollaries. Which one does Chapter 5 need?
Check your understanding
Two functions have the same derivative on an interval. What follows?
Answer: A
Why: Their difference has zero derivative, so by the first corollary it is constant.
Real world
A motorway uses average-speed cameras. A vehicle is recorded entering a 12-mile stretch at 14:00:00 and leaving at 14:10:00. The speed limit throughout is 60 mph, and the driver argues that no camera ever measured their instantaneous speed.
Discussion prompt
Evaluate the argument mathematically. What exactly can be asserted, and what cannot?
Hint: Position is a continuous, differentiable function of time.
Answer:
\[ \text{average speed} = \frac{12 \text{ mi}}{\tfrac{1}{6}\text{ h}} = 72 \text{ mph} \]
Position is a continuous function of time and differentiable throughout — a vehicle cannot teleport, and its velocity exists at every instant. So the Mean Value Theorem applies, and there is some instant during those ten minutes at which the speedometer read exactly 72 mph.
The driver's argument fails. No instantaneous measurement was needed: two position-and-time records force the conclusion mathematically, and this is precisely why average-speed enforcement is legally sound in a way that eyewitness estimates are not.
What cannot be asserted is equally definite. The theorem gives no information about WHEN the 72 mph occurred, how long it was sustained, or what the maximum speed was — which might have been far higher. It also cannot rule out the vehicle having been stationary for part of the stretch, since a long stop simply forces a higher speed elsewhere.
That combination — an unassailable existence claim paired with total silence about location — is the signature of every existence theorem in this course, and it is what makes the Mean Value Theorem both powerful and limited.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is 'a positive derivative means an increasing function' not obvious?
Correct: Because it connects instantaneous information to interval behaviour, which needs the theorem.
\[ f(x_{2})-f(x_{1}) = f'(c)(x_{2}-x_{1}): \text{ one point does all the work} \]
Why: Knowing the slope at every individual point does not immediately tell you anything about the values at two separated points — there is a genuine logical gap. The Mean Value Theorem bridges it by producing a single intermediate point where the average rate is attained, and the sign of the derivative there settles the inequality. The picture is compelling long before the proof exists, which is exactly why the theorem is easy to undervalue.
Explain it
They think the Mean Value Theorem is a technicality with no content.
Discussion prompt
In four sentences or fewer, give them a case where it does real work.
Hint: Use the speed cameras.
Answer:
Ask them how a camera pair twelve miles apart can prove a car exceeded 60 mph when neither camera measures speed. The average works out at 72 mph, and the theorem says some instant during the journey had exactly that speedometer reading — so the offence is certain even though nobody observed it.
That is a conclusion about an unobserved instant, derived from two position measurements and nothing else. Whatever else the theorem is, it is not a technicality: it is what lets you convert what you can measure into what you actually want to know.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For hypotheses, write closed for continuity and open for differentiability and check them separately. For finding c, compute the chord's slope first and then solve the derivative equation. For the proof, remember it is one subtraction that makes the endpoints level. For corollaries, apply the theorem between two arbitrary points and read the signs of the two factors. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a curve with equal endpoint values, mark the horizontal tangent, and write Rolle's theorem beneath with its three hypotheses listed. Below, draw a second curve with unequal endpoints, draw the chord and a parallel tangent, and write the Mean Value Theorem with the chord's slope labelled as the average rate. Beside it, draw the tilting picture: the same curve with a sloping chord, an arrow, and the difference function with level endpoints. In the middle of the page, write the proof in five lines from the auxiliary function to the conclusion. In the lower half, draw the two counterexamples — the absolute value with a level chord and no level tangent, and a step function — labelling which hypothesis each violates. At the bottom, write the four corollaries in a two-column table of condition and consequence, and beneath them write the five-line argument that a positive derivative makes a function increasing. In a margin, write the sign function and one sentence on why the corollaries need a connected interval.
If your increasing-function argument uses the derivative at more than one point, look again — the whole leverage of the theorem is that a single unspecified c does all the work.
Recap
Five things, and the last four sections of this chapter all rest on the fifth.
| If you see | Then |
|---|---|
| Equal endpoint values | Rolle gives a horizontal tangent inside |
| Unequal endpoints | Some tangent is parallel to the chord |
| A corner inside the interval | The theorem does not apply |
| A vertical tangent at an endpoint | The theorem still applies |
| f' = 0 on an interval | The function is constant there |
| f' = g' on an interval | They differ by a constant |
| A disconnected domain | The corollaries can fail |
Section 4.5 now has its licence. Everything it says about where a function rises, falls and turns follows from the corollaries proved here, and the second derivative adds a second layer of the same reasoning about how the graph bends.
OpenStax Calculus Volume 1, §4.4 The Mean Value Theorem §4.4, pp. 329-338 — everything on these slides traces back here
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