Using the Mean Value Theorem: Study Guide (Calculus AB 5.1)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

[a, b] is the closed interval a ≤ x ≤ b, endpoints included. (a, b) is the open interval a < x < b, endpoints left out. The average rate of change of f on [a, b] is

(f(b) − f(a))/(b − a)

which is the gradient (slope) of the secant line joining (a, f(a)) and (b, f(b)). We use MVT as short for the Mean Value Theorem, but in a written answer name the theorem in full at least once.

The idea: your average speed was your actual speed at some moment

A fictional train covers 36 km in 30 minutes. Its average speed is 36/30 = 1.2 km per minute, which is 72 km/h. The train speeds up, slows down and maybe stops at a signal. Can you be sure that at some instant its speedometer read exactly 72 km/h?

Yes. If the speed was always below 72 km/h, the train would cover less than 36 km in 30 minutes. If it was always above, it would cover more. So the speed must be at or above 72 km/h at some times and at or below it at others, and a smoothly changing speed must hit 72 km/h exactly at some instant. You do not know when. You only know it happened.

That is the Mean Value Theorem. Like the Intermediate Value Theorem from Topic 1.16, it is an existence theorem: it tells you that something happens inside an interval without telling you where.

The theorem, part by part

Mean Value Theorem. Suppose f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there is at least one number c in (a, b) with f′(c) = (f(b) − f(a))/(b − a).

The left side is an instantaneous rate (the gradient of a tangent line). The right side is an average rate (the gradient of the secant line). Geometrically: somewhere strictly between a and b, the tangent line is parallel to the secant line.

Part	What it says	How you check it
Condition 1	f is continuous on [a, b]	Name the function type, use a given fact, or check each join of a piecewise function
Condition 2	f is differentiable on (a, b)	Look for corners, cusps, vertical tangents and breaks inside the interval
Conclusion	f′(c) = average rate for at least one c in (a, b)	Calculate the average rate with numbers, then name the theorem

Three details matter.

• Closed for continuity, open for differentiability. The function must join up at the endpoints, but it does not need a derivative there. For example, √x is continuous on [0, 4] and differentiable on (0, 4), even though it has no derivative at 0.
• “At least one”. There may be several suitable values of c. The theorem never says “exactly one”.
• c is strictly inside. A solution of f′(x) = average rate that lies at an endpoint or outside [a, b] does not count.

A special case (Rolle’s theorem). If f(a) = f(b), the average rate is 0, so there is a c in (a, b) with f′(c) = 0: a horizontal tangent. For example, r(x) = x² − 4x + 7 has r(1) = r(3) = 4, and r′(2) = 0 with 2 in (1, 3). You do not need the name; it is simply the Mean Value Theorem with a zero average rate.

Writing a justification

Free-response questions usually say “justify” or “explain why there must be a value…”. A complete answer has three parts.

1. Both conditions, with a reason. “f is a polynomial, so it is continuous on [a, b] and differentiable on (a, b).” Or: “f is given as differentiable, so it is also continuous.” Differentiability implies continuity (Topic 2.4), so one sentence can cover both.
2. The average rate, with numbers. “(f(b) − f(a))/(b − a) = … .”
3. The conclusion, naming the theorem. “By the Mean Value Theorem, there is at least one c in (a, b) with f′(c) = … .”

Leaving out the conditions, or writing “by MVT” without the numbers, loses the justification.

Tables of values. You cannot check a formula, so the question must tell you the function is differentiable (or twice differentiable). Then compute the average rate on each pair of table values. The theorem applies to any sub-interval [p, q] of the domain, not only to neighbouring columns. Choose the sub-interval whose average rate matches the value you are asked about.

Worked example 1: finding c for a polynomial

Question. Let f(x) = x³ − 6x + 1. Show that f satisfies the conditions of the Mean Value Theorem on [0, 3], and find every value of c that the theorem guarantees.

1. Conditions. f is a polynomial, so it is continuous on [0, 3] and differentiable on (0, 3).
2. End values. f(0) = 1. f(3) = 27 − 18 + 1 = 10.
3. Average rate. (10 − 1)/(3 − 0) = 9/3 = 3.
4. Set the derivative equal to it. f′(x) = 3x² − 6. Solve 3x² − 6 = 3, so 3x² = 9, x² = 3, x = ±√3.
5. Keep only values in (0, 3). −√3 is negative, so reject it. √3 ≈ 1.732 lies in (0, 3).

Answer. c = √3. Figure 1 shows the tangent there.

Check. f′(√3) = 3(3) − 6 = 3, which equals the average rate.

Notice. c is not the midpoint 1.5: f′(1.5) = 3(2.25) − 6 = 0.75, not 3. Nor is the average rate the average of f′(0) = −6 and f′(3) = 21, which is 7.5. The theorem compares f′(c) with the secant gradient, nothing else.

Worked example 2: a table in context

Question. A fictional train moves along a straight track. Its distance from the station is s(t) km, t minutes after departure. s is differentiable. Some values are shown.

t (minutes)	0	4	10	15	20
s(t) (km)	0	3.2	9.8	14.8	16.0

(a) Must there be a time t in (4, 10) at which the train’s velocity is 1.1 km per minute? Justify. (b) Must there be a time t in (10, 20) at which the velocity is 1.0 km per minute? Justify. (c) Does the Mean Value Theorem guarantee a time in (10, 20) at which the velocity is 1.3 km per minute?

(a) s is differentiable, so it is continuous on [4, 10] and differentiable on (4, 10). The average rate is

(s(10) − s(4))/(10 − 4) = (9.8 − 3.2)/6 = 6.6/6 = 1.1 km per minute.

By the Mean Value Theorem there is a time t in (4, 10) with s′(t) = 1.1. Since s′ is the velocity, yes: at some time between 4 and 10 minutes the train moves at 1.1 km per minute (66 km/h).

(b) On the whole interval [10, 20] the average rate is (16.0 − 9.8)/10 = 0.62, which is not 1.0. Do not stop there. Try a sub-interval: on [10, 15], (14.8 − 9.8)/5 = 5.0/5 = 1.0. s is continuous on [10, 15] and differentiable on (10, 15), so by the Mean Value Theorem there is a time in (10, 15), and therefore in (10, 20), with velocity 1.0 km per minute. Yes.

(c) The average rates from the table inside [10, 20] are 1.0, 0.24 and 0.62. None is 1.3, so the theorem gives no guarantee. The train may or may not have reached 1.3 km per minute; the table does not say. “No guarantee” is not the same as “impossible”.

Worked example 3: when a condition fails

Question. Let f(x) = x^(2/3), the square of the cube root of x, on [−1, 8]. Is there a c in (−1, 8) with f′(c) equal to the average rate? Explain the result.

1. End values. f(−1) = ((−1)^(1/3))² = (−1)² = 1. f(8) = 2² = 4.
2. Average rate. (4 − 1)/(8 − (−1)) = 3/9 = 1/3.
3. Derivative. For x ≠ 0, f′(x) = (2/3)x^(−1/3) = 2/(3 · x^(1/3)). At x = 0 this is undefined: the graph has a cusp (a sharp point with a vertical tangent).
4. Solve f′(x) = 1/3. For x > 0: 2/(3x^(1/3)) = 1/3 gives x^(1/3) = 2, so x = 8. For x < 0: x^(1/3) is negative, so f′(x) is negative and cannot equal 1/3.
5. Check the interval. The only solution is x = 8, an endpoint. There is no c in (−1, 8).

Explanation. This does not contradict the theorem. f is continuous on [−1, 8], but it is not differentiable at x = 0, which lies inside the interval. Condition 2 fails, so the theorem never promised anything.

Other ways a condition fails. A corner (as in |x − k| at x = k), a vertical tangent, a jump, a hole or an asymptote inside the interval each break a condition. A function such as 1/x on [−1, 1] is not even continuous, because it is undefined at 0.

No promise is not the same as impossible. A function can fail a condition and still have a suitable c by chance. The theorem simply stays silent.

Common misconceptions

• “c is the midpoint of [a, b].” Only by coincidence. In Worked example 1, the midpoint is 1.5 but c = √3.
• “The average rate is the average of f′(a) and f′(b).” It is (f(b) − f(a))/(b − a), the secant gradient.
• “There is exactly one c.” The theorem says at least one. A wiggly graph can have many.
• Keeping solutions outside (a, b). Always test each solution against the open interval.
• Forgetting the chain rule when you differentiate before solving f′(c) = average rate.
• Skipping the conditions. “By MVT, f′(c) = 3” earns nothing without continuity and differentiability, each with a reason.
• Using only neighbouring table columns. Any sub-interval [p, q] in the table is allowed.
• Confusing the Mean Value Theorem with the Intermediate Value Theorem. The IVT is about values of f (needs only continuity). The MVT is about values of f′ (needs differentiability too).
• “The condition fails, so there is no such c.” A failed condition only removes the guarantee.

Where this leads

The Mean Value Theorem is the reason that the sign of f′ controls whether f increases or decreases, which you will use from Topic 5.3 onwards. It is also behind the link between derivatives and accumulation in Unit 6. Next, Topic 5.2, Extreme Value Theorem, Global Versus Local Extrema, and Critical Points, introduces a second existence theorem and the critical points where extreme values can occur. The previous topic was Topic 4.7, L’Hospital’s Rule. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.