- The intermediate value theorem (IVT) says a continuous function on \( [a, b] \) takes every value between \( f(a) \) and \( f(b) \). It is used to prove that a root or a target value exists.
- The mean value theorem (MVT) says a smooth function on \( [a, b] \) has at least one point where the instantaneous rate of change equals the average rate of change.
- Both are existence theorems: they guarantee that a point exists, but they do not tell you where it is.
- In our case, a car that covers 12 km between two cameras in 6 minutes averages 120 km/h, so by the MVT it was doing exactly 120 km/h at some instant, even if no camera saw it.
- Exam point: before quoting either theorem, state its conditions (continuous on the closed interval; for the MVT, also differentiable on the open interval).
The case: the speed camera that never saw you speeding
A motorway has a speed limit of 100 km/h. An average-speed check uses two cameras 12 km apart. Each camera reads a car’s number plate and records the time. A driver passes the first camera at 95 km/h, comfortably under the limit. She passes the second camera exactly 6 minutes later, again at a modest speed. Neither camera ever saw her over 100 km/h.
Her average speed was
She argues: “You cannot prove I was ever doing more than 100. You only measured me at two points, and I was under the limit at both.” Is she right? Does an average of 120 km/h prove that at some actual moment her speedometer read 120?
A second question follows. She entered the zone at 95 km/h. If at some moment she was doing 120, must she also have passed through exactly 100 km/h at some moment? Common sense says yes, but we want a reason that holds for every possible journey, not just a typical one. Two theorems answer these questions: the mean value theorem and the intermediate value theorem. (The figures are illustrative.)
The concept: two existence theorems
Most of calculus is about finding things: a derivative, a maximum, an area. These two theorems are different. They are about proving that something exists. They are the logical foundation underneath many results you use without thinking, such as “if the derivative is positive, the function is increasing”.
Start with the IVT, because it is the simpler idea. Imagine drawing the graph of a continuous function without lifting your pen. If you start below a horizontal line and finish above it, your pen must cross the line somewhere. That is all the IVT says. The key word is continuous: no jumps, no gaps, no asymptotes in the interval. A function with a jump could leap over the line without touching it.
Here \( a \) and \( b \) are the ends of the interval, and \( f(a) \), \( f(b) \) are the function values there. The right-hand side, \( \frac{f(b) – f(a)}{b – a} \), is the average rate of change: the gradient of the straight line (the secant, or chord) joining the two end points. \( f'(c) \) is the instantaneous rate of change at the point \( c \): the gradient of the tangent there. The theorem says some tangent is parallel to the chord. The number \( c \) is strictly between \( a \) and \( b \), and there may be more than one such \( c \) [1].
Mathematics Made Simple explains the mean value theorem and IVT with worked examples and graphs.Get the textbook · $12
Why is the MVT true? Picture the chord and slide it upwards, keeping it parallel, until it last touches the curve. At that point of touching, the chord is a tangent, so the curve’s gradient equals the chord’s gradient. The formal proof uses a special case called Rolle’s theorem: if \( f(a) = f(b) \), then somewhere \( f'(c) = 0 \). In other words, if you finish at the same height you started, at some point you were moving neither up nor down.
The two theorems need different conditions. The IVT needs only continuity. The MVT also needs differentiability, meaning no corners or cusps between \( a \) and \( b \). A position-time graph for a real car is smooth, because a car cannot change its speed instantly. So both theorems apply to our driver.
Why this method fits the case
The cameras give us exactly two data points: position at two times. We do not know the speed at any moment in between. This is the situation where existence theorems shine. We cannot compute the speed at each instant, but we can still prove something about it.
Let \( s(t) \) be the distance travelled, in km, \( t \) hours after the first camera. Then \( s(0) = 0 \) and \( s(0.1) = 12 \). The car’s speed is \( s'(t) \). The MVT applies because \( s \) is continuous and differentiable. So there is a time \( c \) between 0 and 0.1 hours with \( s'(c) = 120 \). Her speedometer really did read 120 km/h at some instant.
For the second question, apply the IVT to the speed itself. The speed \( v(t) = s'(t) \) is continuous (a car’s speed does not jump). It was 95 at \( t = 0 \) and 120 at \( t = c \). Since 100 lies between 95 and 120, the IVT guarantees a moment between 0 and \( c \) when the speed was exactly 100 km/h.
When are these theorems not the right tool?
- When you need to know where. They do not tell you when she hit 120 km/h. To find the time you need the actual speed function, or more data.
- When the conditions fail. \( f(x) = \frac{1}{x} \) on \( [-1, 1] \) goes from \(-1\) to 1 but never equals 0, because it is not continuous at 0. \( f(x) = |x| \) on \( [-1, 1] \) has average gradient 0 but its derivative is never 0, because it has a corner at 0.
- When you need the maximum. The MVT shows the speed reached at least 120. It does not give the top speed. For that you need the methods of extreme values and optimisation.
Setting it up
To see the theorems in action, we use one possible speed profile that fits the camera data. It is not the only one, which is the point: the conclusions hold for every smooth profile.
| Quantity | Symbol | Value or formula |
|---|---|---|
| Distance between cameras | \( s(6) – s(0) \) | 12 km |
| Time between cameras | \( b – a \) | 6 minutes = 0.1 hours |
| Average speed | \( \frac{s(b) – s(a)}{b – a} \) | 120 km/h |
| Speed limit | \( L \) | 100 km/h |
| Illustrative speed profile (\( t \) in minutes) | \( v(t) \) | \( 95 + 12.5\pi \sin\left(\frac{\pi t}{6}\right) \) km/h |
| Speed at each camera | \( v(0), v(6) \) | 95 km/h |
| Peak speed in the profile | \( v(3) \) | \( 95 + 12.5\pi \approx 134.3 \) km/h |
The amplitude \( 12.5\pi \approx 39.27 \) was chosen so that the profile covers exactly 12 km in 6 minutes. You can check this by integration: \( \frac{1}{60}\int_0^6 v(t)\,dt = \frac{1}{60}\left(570 + 12.5\pi \cdot \frac{12}{\pi}\right) = \frac{720}{60} = 12 \).
Solving it
Use the MVT to prove the car reached 120 km/h, then find when it did so for the illustrative profile.
The car must have been doing 120 km/h at least once; in this profile it happens twice, at about 1.32 and 4.68 minutes.
Find the value of \( c \) guaranteed by the MVT for \( f(x) = x^3 \) on \( [0, 2] \).
The tangent at \( x = \frac{2}{\sqrt3} \approx 1.155 \) is parallel to the chord from \( (0, 0) \) to \( (2, 8) \).
Use the IVT to show that \( x^3 + x – 5 = 0 \) has a root between 1 and 2, then narrow it to an interval of width 0.1.
The sign change shows a root lies in \( (1.5, 1.6) \); the same idea drives the decimal-search and bisection methods.
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What the answer tells us
The driver’s argument fails. An average speed of 120 km/h is a mathematical guarantee that she was doing 120 km/h at some instant. It does not matter that no camera caught her at that moment, and it does not matter how she varied her speed. If she had stayed at or below 100 km/h the whole way, she could have covered at most \( 100 \times 0.1 = 10 \) km in 6 minutes, not 12. The MVT is the precise form of that common-sense argument.
The IVT adds a second layer. Her speed started at 95 km/h and, by the MVT, reached 120 km/h. Speed is continuous, so it passed through every value in between, including exactly 100 km/h. For our profile that happens at about 0.24 minutes, roughly 15 seconds after the first camera. Figure 1 shows the whole picture: the speed curve starts and ends below the limit, crosses 100, and crosses the 120 average line twice.
Is this reasonable? Yes. The profile peaks at 134.3 km/h, which is well above the average, and spends some time below it near each camera. That balance is what the MVT forces: if speed was below average for part of the trip, it must be above average for another part. Notice too that the guaranteed speed is a minimum claim about the peak. The true top speed in our profile, 134 km/h, is higher than 120. The MVT only promises that the speed hit the average; the real maximum can be much higher.
How sensitive is the conclusion? It depends only on two numbers: the 12 km and the 6 minutes. If the cameras’ clocks were out by 30 seconds, so the true time was 6.5 minutes, the average would be \( \frac{12}{6.5} \times 60 \approx 110.8 \) km/h. That is still above the limit. The conclusion breaks only if the true time is above 7.2 minutes, because \( \frac{12}{7.2} \times 60 = 100 \). This is why average-speed checks depend on accurately synchronised clocks and a precisely known distance.
For the theory behind continuity, differentiability and these existence theorems, see the Mathematics Made Simple textbook, and practise table-based “must there be…?” questions in the Mathematics Made Simple Practice Questions book.
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Practice questions
State the theorem you use and check its conditions each time.
EasyQ1. Show that \( \cos x = x \) has a solution between 0 and 1 (radians). Show worked answer
\( g(0) = 1 > 0 \); \( g(1) = \cos 1 – 1 \approx -0.460 < 0 \).
By the IVT, \( g(c) = 0 \) for some \( c \) in \( (0, 1) \), so \( \cos c = c \).
EasyQ2. Find the value of \( c \) given by the MVT for \( f(x) = x^2 \) on \( [1, 5] \). Show worked answer
\( f'(c) = 2c = 6 \Rightarrow c = 3 \), which is in \( (1, 5) \).
MediumQ3. Find all values of \( c \) given by the MVT for \( f(x) = x^3 – 6x^2 + 9x + 1 \) on \( [0, 4] \). Show worked answer
\( f'(c) = 3c^2 – 12c + 9 = 1 \Rightarrow 3c^2 – 12c + 8 = 0 \).
\( c = \frac{12 \pm \sqrt{144 – 96}}{6} = 2 \pm \frac{2\sqrt{3}}{3} \approx 0.845 \) or \( 3.155 \). Both lie in \( (0, 4) \).
MediumQ4. For \( f(x) = |x| \) on \( [-1, 1] \), show that no \( c \) satisfies the MVT conclusion, and explain why this does not contradict the theorem. Show worked answer
But \( f'(x) = -1 \) for \( x < 0 \) and \( 1 \) for \( x > 0 \), never 0.
No contradiction: \( |x| \) is not differentiable at \( x = 0 \), so the MVT’s conditions are not met.
ExamQ5. \( f \) is twice differentiable with \( f(0) = 4 \), \( f(2) = 10 \), \( f(5) = 1 \), \( f(8) = 7 \). (a) What is the least number of solutions of \( f(x) = 5 \) in \( (0, 8) \)? Justify. (b) Must there be \( c \) in \( (2, 5) \) with \( f'(c) = -3 \)? Justify. (c) Explain why \( f'(x) = 0 \) has at least two solutions in \( (0, 8) \). [6] Show worked answer
(b) \( f \) is differentiable, and \( \frac{f(5) – f(2)}{5 – 2} = \frac{1 – 10}{3} = -3 \), so yes, by the MVT. [2]
(c) By the MVT, \( f’ \) takes the value 3 in \( (0, 2) \), \(-3\) in \( (2, 5) \) and 2 in \( (5, 8) \). \( f’ \) is continuous (as \( f \) is twice differentiable), so by the IVT \( f’ = 0 \) between the first two points and again between the last two. [2]
Frequently asked questions
What is the difference between the IVT and the MVT?
The IVT is about the values of a function: a continuous function hits every value between its end values. The MVT is about the gradient: a smooth function has a point where its gradient equals its average gradient. The IVT needs continuity; the MVT also needs differentiability.
Is the extreme value theorem related?
Yes. The extreme value theorem says a continuous function on a closed interval \( [a, b] \) has a maximum and a minimum. It is the third existence theorem in this family, and it is used to prove Rolle’s theorem and so the MVT. Guide step 16 covers it alongside the MVT.
Can the value \( c \) be at an end point?
No. Both theorems guarantee \( c \) strictly inside the interval, in \( (a, b) \). In the IVT, if \( N \) equals \( f(a) \) or \( f(b) \) the result is trivial, so the theorem is stated for values strictly between.
Where is the MVT used later in calculus?
It proves that a function with positive derivative is increasing, that two functions with the same derivative differ by a constant (the basis of “+ C” in integration), and it underlies error bounds for Taylor polynomials and the fundamental theorem of calculus.
References
- Strang, G. and Herman, E. (2016) Calculus Volume 1. Houston, TX: OpenStax. Available at: https://openstax.org/details/books/calculus-volume-1.
- College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description. New York: College Board.
- Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment. Available at: cambridgeinternational.org.
- International Baccalaureate Organization (2019) Mathematics: analysis and approaches guide (first assessment 2021). Cardiff: IBO.
