The mean value theorem
Two cameras time a car over a known stretch of highway and find an average speed above the limit. Without ever seeing the car speed, can they be sure its speedometer once showed that average?
Where you are 11 in Applications of derivatives
Why this exists
A derivative is a local fact: f'(c) describes f in a tiny window around c. Most questions are about a whole interval. If the derivative is positive everywhere, does the function rise from one end to the other? If two functions have the same derivative, how different can they be? If a speed never passes 65 miles per hour, how far can a car get in ten minutes? None of these follows from the definition of the derivative at a point alone. Something has to connect the slope at single points to the change over an interval.
Biljana drives through a highway speed check that works this way. One camera records the moment her car passes, and a second camera a known distance down the road records it again. The authority divides distance by time and fines any average speed above the limit. Goce, a passenger, objects that the cameras never measured her speed at any instant. The mean value theorem answers him: a car whose average speed over the stretch is 75 miles per hour was traveling at exactly 75 miles per hour at some instant in between.
Definition
The lesson rests on two theorems already met. The The extreme value theorem says a function continuous on [a, b] reaches an absolute maximum and an absolute minimum there. Fermat’s theorem (Critical points and local versus absolute extrema) says that if f has a local maximum or minimum at a point c inside an interval and f'(c) exists, then f'(c) = 0.
Rolle’s theorem. Suppose f is
- continuous on the closed interval [a, b],
- differentiable on the open interval (a, b), and
- f(a) = f(b).
Then there is at least one number c in (a, b) with f'(c) = 0.
The mean value theorem. Suppose f is continuous on [a, b] and differentiable on (a, b). Then there is at least one number c in (a, b) with
The right side is the slope of the secant line through (a, f(a)) and (b, f(b)), the average rate of change of f over [a, b]. Rolle’s theorem is the case f(a) = f(b), where that slope is 0.
Intuition
A smooth graph that starts and ends at the same height must turn around somewhere, and where it turns, its tangent is level. Tilt the picture and the same sentence becomes the mean value theorem: a smooth graph from one point to another must, somewhere in between, run exactly parallel to the chord joining them (Figure 1).
In the language of motion, f(b) - f(a) over b - a is an average velocity and f'(c) is an instantaneous velocity. The theorem says the average is always one of the instantaneous values. The same idea is why a difference quotient taken from a table, as in The derivative as a function, equals the derivative at some point inside the interval.
Compare the intermediate value theorem of The intermediate value theorem. That theorem is about the values of f: a continuous function takes every height between f(a) and f(b). The mean value theorem is about the values of f': one particular slope, the average one, is taken. Both say only that a number exists, not where.
Each hypothesis of Rolle’s theorem matters. Figure 2 drops one at a time, and in each picture no level tangent appears.
Results
Day 1: the two theorems
Rolle’s theorem. If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then f'(c) = 0 for some c in (a, b).
Mean value theorem. If f is continuous on [a, b] and differentiable on (a, b), then f(b) - f(a) = f'(c)(b - a) for some c in (a, b). Geometrically, the tangent at c is parallel to the secant.
Every hypothesis is needed.
| Hypothesis dropped | Function on [0, 2] | What goes wrong |
|---|---|---|
| differentiable on (0, 2) | \lvert x - 1 \rvert | slopes are -1 and 1, never 0 |
| continuous on [0, 2] | x for x < 2, value 0 at x = 2 | slope is always 1 |
| f(a) = f(b) | x^2 | slope 2x is 0 only at the end x = 0 |
Day 2: consequences
Throughout, I is an interval and “inside I” means not at an end.
Zero derivative. If f'(x) = 0 at every point inside I and f is continuous on I, then f is constant on I.
Equal derivatives. If f'(x) = g'(x) at every point inside I and both are continuous on I, then f(x) = g(x) + C on I for some constant C.
Sign of the derivative. If f is continuous on I and f'(x) > 0 at every point inside I, then f is increasing on I: x_1 < x_2 gives f(x_1) < f(x_2). If f'(x) < 0 inside I, then f is decreasing on I.
Bounded change. If f is continuous on [a, b] and \lvert f'(x) \rvert \le M on (a, b), then
Why it is true
Rolle’s theorem. Since f is continuous on [a, b], the extreme value theorem gives an absolute maximum value M and an absolute minimum value m. If both are reached only at the ends, then M = m = f(a) = f(b), so f is constant and f'(c) = 0 for every c in (a, b). Otherwise at least one of them is reached at a point c inside (a, b). An absolute extreme at an inside point is also a local extreme, and f'(c) exists because f is differentiable on (a, b). By Fermat’s theorem, f'(c) = 0.
The mean value theorem, by tilting. Let m = \dfrac{f(b) - f(a)}{b - a}, and let L(x) = f(a) + m(x - a) be the secant line. Define the auxiliary function
the vertical gap between the curve and the chord. Then g is continuous on [a, b] and differentiable on (a, b), because f and the line L are. At the ends, g(a) = f(a) - f(a) = 0 and g(b) = f(b) - f(a) - m(b - a) = 0. Rolle’s theorem gives c in (a, b) with g'(c) = 0. Since g'(x) = f'(x) - m, this says f'(c) = m. Subtracting the chord levels it, and Rolle’s theorem does the rest.
Zero derivative. Take any x_1 < x_2 in I. The mean value theorem on [x_1, x_2] gives f(x_2) - f(x_1) = f'(c)(x_2 - x_1) with c between them, so inside I. Since f'(c) = 0, f(x_2) = f(x_1). Any two values are equal, so f is constant.
Equal derivatives. Apply the previous result to h = f - g, whose derivative is f' - g' = 0 inside I. Then h is a constant C, so f = g + C. This is where the +\,C of antiderivatives comes from: once one antiderivative is known on an interval, every other differs from it by a constant.
Sign of the derivative. For x_1 < x_2 in I, f(x_2) - f(x_1) = f'(c)(x_2 - x_1). Both factors on the right are positive, so f(x_2) > f(x_1). With f' < 0 the product is negative.
Bounded change. \lvert f(b) - f(a) \rvert = \lvert f'(c) \rvert (b - a) \le M(b - a).
Worked examples
Day 1: Biljana checks Rolle’s hypotheses
Show that f(x) = 9x - x^3 satisfies the hypotheses of Rolle’s theorem on [0, 3], and find every c the theorem promises.
Solutionf is a polynomial, so it is continuous and differentiable everywhere, in particular on [0, 3] and (0, 3). At the ends, f(0) = 0 and f(3) = 27 - 27 = 0, so f(0) = f(3). All three hypotheses hold.
To find c, solve f'(x) = 9 - 3x^2 = 0, which gives x^2 = 3 and \displaystyle x = \pm\sqrt{3}. Only \displaystyle \sqrt{3} \approx 1.732 lies in (0, 3), so \displaystyle c = \sqrt{3}. There \displaystyle f(\sqrt{3}) = 9\sqrt{3} - 3\sqrt{3} = 6\sqrt{3}, the top of the hump in Figure 1, where the tangent is horizontal.
Day 1: three functions that miss a hypothesis
For each function, decide which hypothesis of Rolle’s theorem fails on [0, 2], and whether f'(c) = 0 for some c in (0, 2) anyway. (a) f(x) = \lvert x - 1 \rvert. (b) g(x) = x for 0 \le x < 2, with g(2) = 0. (c) h(x) = x^2.
Solution(a) f(0) = 1 = f(2) and f is continuous, but it has a corner at x = 1, where it is not differentiable. Its slope is -1 on (0, 1) and 1 on (1, 2), never 0. The proof breaks at the maximum-or-minimum step: the minimum sits at x = 1, but Fermat’s theorem needs f'(1) to exist.
(b) g(0) = 0 = g(2) and g'(x) = 1 on all of (0, 2), but g is not continuous at 2, since \lim_{x \to 2^-} g(x) = 2 \ne g(2). No horizontal tangent. Here the extreme value theorem fails: g has no maximum on [0, 2].
(c) h is continuous and differentiable, but h(0) = 0 and h(2) = 4. The only solution of h'(x) = 2x = 0 is x = 0, an end, so no c lies in (0, 2). Figure 2 shows all three.
Day 1: finding the point the mean value theorem promises
Verify the hypotheses of the mean value theorem and find every c it promises. (a) \displaystyle f(x) = \sqrt{x} on [1, 9]. (b) p(x) = x^3 - 4x on [-2, 4].
Solution(a) \displaystyle \sqrt{x} is continuous on [0, \infty) and differentiable on (0, \infty), so on [1, 9] and (1, 9). The secant slope is
Set \displaystyle f'(c) = \dfrac{1}{2\sqrt{c}} = \dfrac{1}{4}: then \displaystyle \sqrt{c} = 2 and c = 4, which lies in (1, 9). The tangent at (4, 2) is the line parallel to the secant in Figure 1.
(b) A polynomial satisfies both hypotheses. Here p(-2) = -8 + 8 = 0 and p(4) = 64 - 16 = 48, so the secant slope is \dfrac{48 - 0}{4 - (-2)} = 8. Solve p'(c) = 3c^2 - 4 = 8: c^2 = 4, so c = 2 or c = -2. The theorem places c in the open interval (-2, 4), so c = -2, the left end, does not count, and the answer is c = 2. (At x = -2 the tangent happens to be parallel to the secant as well, but it touches at an end.)
Day 1: the point of the arc farthest from its chord
Let f(x) = e^x on [0, 1]. Find the c given by the mean value theorem, and show that at c the vertical gap between the secant and the curve is largest.
SolutionThe secant slope is \dfrac{e^1 - e^0}{1 - 0} = e - 1. Solving f'(c) = e^c = e - 1 gives c = \ln(e - 1) \approx 0.541, which lies in (0, 1) because 1 < e - 1 < e.
The secant line is L(x) = 1 + (e - 1)x, and the gap is the auxiliary function of the proof, up to sign: d(x) = L(x) - e^x. It is 0 at both ends and positive inside, since the curve bends below its chord. By the extreme value theorem d has a maximum on [0, 1], inside the interval because the ends give 0. There d'(x) = (e - 1) - e^x = 0, which is exactly the equation for c. The largest gap is
So the tangent parallel to the chord touches the curve at the point farthest, vertically, from the chord. This is the picture inside the proof: Rolle’s theorem finds the turning point of the gap.
Day 2: Zlatko’s two antiderivatives
Zlatko and Biljana each look for a function whose derivative is \dfrac{1}{(x + 1)^2} on (-1, \infty). Zlatko offers F(x) = \dfrac{x}{x + 1} and Biljana offers G(x) = -\dfrac{1}{x + 1}. Show that both are right, explain how they can differ, and find the one function with this derivative whose value at 0 is 3.
SolutionBy the quotient rule, F'(x) = \dfrac{(x + 1) - x}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}, and G'(x) = \dfrac{1}{(x + 1)^2} by the power rule on -(x + 1)^{-1}. The derivatives agree on the interval (-1, \infty), so the functions differ by a constant there. Indeed
Any function with this derivative on (-1, \infty) is G(x) + C for one constant C. The condition G(0) + C = -1 + C = 3 gives C = 4, so the function is 4 - \dfrac{1}{x + 1}. The interval matters: on the whole domain x \ne -1, two separate constants could be used on the two sides of -1.
Day 2: a logarithm inequality from the sign of a derivative
Show that \ln(1 + x) < x for every x > -1 except x = 0, where the two are equal.
SolutionThis example uses \dfrac{d}{dx}\big[\ln x\big] = \dfrac{1}{x} for x > 0, derived in Derivatives of logarithms and logarithmic differentiation; with the chain rule it gives \dfrac{d}{dx}\big[\ln(1 + x)\big] = \dfrac{1}{1 + x}.
Let h(x) = x - \ln(1 + x) for x > -1, with h(0) = 0. It is differentiable there, with
The denominator is positive, so h' has the sign of x.
For x > 0, apply the mean value theorem on [0, x]: h(x) - h(0) = h'(c)\,x with 0 < c < x. Here h'(c) > 0 and x > 0, so h(x) > h(0) = 0.
For -1 < x < 0, apply it on [x, 0]: h(0) - h(x) = h'(c)(0 - x) with x < c < 0. Now h'(c) < 0 and -x > 0, so h(0) - h(x) < 0, that is, h(x) > 0.
In both cases x - \ln(1 + x) > 0, so \ln(1 + x) < x. For instance, \ln 2 \approx 0.693 < 1. The theorem turned a sign of h' into a comparison of values; no table of \ln was needed.
Day 2: the two-camera speed check
Biljana passes camera A at 10:02 a.m. and camera B, 10 miles farther along the highway, at 10:10 a.m. The limit is 65 miles per hour. (a) Show that at some instant she was traveling at exactly her average speed, and find it. (b) Zlatko drives the same stretch and never exceeds 65 miles per hour. If he passes A at 10:02 a.m., what is the earliest he can pass B?
Solution(a) Let s(t) be her distance along the road, in miles, after t hours. A real car’s position is continuous, and its velocity s'(t) exists at every instant, so the mean value theorem applies on the time interval. The interval lasts 8 minutes, \dfrac{8}{60} = \dfrac{2}{15} hour, so
for some instant c between the two cameras (Figure 3). She was over the limit at that instant whatever she did at the cameras themselves, and Goce’s objection fails. The theorem promises the average speed, not a top speed: she may have gone faster at other times.
(b) Now 0 \le s'(t) \le 65, so the bounded-change result on his time interval [t_A, t_B] gives
So t_B - t_A \ge \dfrac{10}{65} = \dfrac{2}{13} hour, which is \dfrac{120}{13} \approx 9.23 minutes, a little over 9 minutes 13 seconds. He reaches B at about 10:11 a.m. at the earliest; any earlier time at B would prove a speed above 65.
Boundaries
- Continuity on the closed interval and differentiability on the open one. Drop either and the conclusion can fail (Example 2). Differentiability at the ends is not needed: \displaystyle \sqrt{x} on [0, 4] has no derivative at 0, yet the theorem applies and gives c = 1.
- Existence only. The theorems promise at least one c, possibly several (Exercise 3), and do not say where. When the hypotheses fail the theorems are silent: x^3 - 3x^2 on [-1, 3] has unequal end values -4 and 0 and still has level tangents at 0 and 2.
- The consequences need an interval. \dfrac{\lvert x \rvert}{x} has derivative 0 at every point of its domain and is not constant, because its domain x \ne 0 is two intervals (Exercise 13).
- A positive derivative is enough for increasing, not necessary. x^3 is increasing on (-\infty, \infty) although its derivative is 0 at x = 0. A derivative that is positive except at isolated points still gives an increasing function, by applying the result on each piece. Reading the sign of f' across a whole domain is the work of Increasing and decreasing from the derivative.
- The speed argument needs a smooth position. It holds because a moving car has a velocity at every instant; a quantity that could jump would escape it.
Where this goes next
The sign of the derivative now controls rising and falling, which is what Increasing and decreasing from the derivative and the The first and second derivative tests use to find and classify the Maxima, minima and critical points of a function, and what Concavity and inflection points applies one derivative higher. The equal-derivatives result is the reason an antiderivative is written with +\,C in Antiderivatives and indefinite integrals, and the mean value theorem is the key step in proving the The fundamental theorem of calculus, which makes The second fundamental theorem of calculus possible. Bounding a change by a bound on the derivative returns in the The Taylor remainder and the Lagrange error bound, a mean value theorem for Taylor polynomials.
Exercises
Day 1: Rolle’s theorem
Decide whether Rolle’s theorem applies on the interval. Where it does, find every c it promises; where it does not, say which hypothesis fails and whether f'(c) = 0 inside the interval anyway.
- x^2 - 4x - 3 on [-1, 5]
- \cos(2x) on [0, \pi]
- x^4 - 8x^2 on [-3, 3]
- 1 - x^{2/3} on [-1, 1]
- \tan x on [0, \pi]
Day 1: the mean value theorem
Find every c the mean value theorem promises, or explain why there is none.
- x^3 on [0, 3]
- \dfrac{4}{x} on [1, 4]
- \ln x on [1, e]
- \sin x on \displaystyle \left[0, \frac{\pi}{2}\right], to three decimal places
- \dfrac{1}{x - 2} on [0, 4]
Day 2: functions with the same derivative
- Find the function f with f'(x) = 3x^2 + 2 for every x and f(1) = 4, and explain why there is only one.
- Show that \tan^2 x and \sec^2 x have the same derivative on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), and find the constant by which they differ.
- Let f(x) = \dfrac{\lvert x \rvert}{x} for x \ne 0. Show that f'(x) = 0 at every point of its domain, and explain why f is not constant.
Day 2: rising, falling and bounding a change
- Show that x^3 + x - 5 is increasing on (-\infty, \infty), and conclude that x^3 + x - 5 = 0 has exactly one real solution. Between which two consecutive integers does it lie?
- Show that \displaystyle \sqrt{1 + x} < 1 + \dfrac{x}{2} for every x > 0.
- A function is differentiable everywhere, with f(1) = 4 and \lvert f'(x) \rvert \le 3 for every x. Find the smallest and largest values f(5) could have.
- Use the mean value theorem for \displaystyle \sqrt{x} on [100, 101] to show that \displaystyle 10 + \dfrac{1}{22} < \sqrt{101} < 10 + \dfrac{1}{20}.
- Zlatko passes one camera at 3:00 p.m. and a second camera 4 miles farther on at 3:05 p.m. (a) Show that at some instant he was traveling at exactly 48 miles per hour. (b) If his speed never exceeded 60 miles per hour, what is the earliest he could have passed the second camera?
Reasoning
- Show that for any quadratic f(x) = px^2 + qx + r with p \ne 0, the c of the mean value theorem on [a, b] is always the midpoint \dfrac{a + b}{2}.
- Use Rolle’s theorem to show that x^3 + 4x - 1 = 0 cannot have two real solutions, and the intermediate value theorem to show it has one.
- Let f have a second derivative everywhere, with f(0) = f(1) = f(2) = 0. Show that f''(c) = 0 for some c in (0, 2).
- Biljana and Goce start a race together and finish in a dead heat. Show that at some instant during the race they were running at the same speed.
- Show that \lvert \sin b - \sin a \rvert \le \lvert b - a \rvert for all numbers a and b.
Answers
- Applies: a polynomial, with f(-1) = 2 = f(5). f'(x) = 2x - 4 = 0 gives c = 2.
- Applies: \cos 0 = 1 = \cos 2\pi. f'(x) = -2\sin(2x) = 0 inside (0, \pi) only at 2x = \pi, so \displaystyle c = \frac{\pi}{2}.
- Applies: f(\pm 3) = 81 - 72 = 9. f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2), so c = -2, 0 and 2, three values.
- Does not apply: f(\pm 1) = 0, but \displaystyle f'(x) = -\frac{2}{3}x^{-1/3} does not exist at x = 0 (a cusp). For x \ne 0 the derivative is never 0, so no c exists.
- Does not apply: \tan 0 = \tan \pi = 0, but \tan x is undefined at \displaystyle \frac{\pi}{2}, so it is not continuous on [0, \pi]. Its derivative \sec^2 x \ge 1 is never 0.
- Secant slope \displaystyle \frac{27 - 0}{3} = 9. 3c^2 = 9 gives \displaystyle c = \sqrt{3} (the root \displaystyle -\sqrt{3} is outside).
- Secant slope \displaystyle \frac{1 - 4}{3} = -1. \displaystyle -\frac{4}{c^2} = -1 gives c = 2.
- Secant slope \displaystyle \frac{1 - 0}{e - 1} = \frac{1}{e - 1}. With \displaystyle (\ln x)' = \frac{1}{x} (Derivatives of logarithms and logarithmic differentiation), \displaystyle \frac{1}{c} = \frac{1}{e - 1} gives c = e - 1 \approx 1.718.
- Secant slope \displaystyle \frac{1 - 0}{\pi/2} = \frac{2}{\pi}. \displaystyle \cos c = \frac{2}{\pi} gives \displaystyle c = \arccos\frac{2}{\pi} \approx 0.881.
- None. Secant slope \displaystyle \frac{\frac{1}{2} - \left(-\frac{1}{2}\right)}{4} = \frac{1}{4}, but \displaystyle f'(x) = -\frac{1}{(x - 2)^2} < 0. No contradiction: f is not continuous at 2.
- f(x) = x^3 + 2x + 1: x^3 + 2x has the right derivative, so every answer is x^3 + 2x + C, and 1 + 2 + C = 4 gives C = 1. Two such functions have equal derivatives on (-\infty, \infty), so they differ by a constant, and agreeing at 1 makes it 0.
- By the chain rule both derivatives are 2\tan x \sec^2 x (for \sec^2 x, use \displaystyle \frac{d}{dx}\big[\sec x\big] = \sec x \tan x). They differ by 1: \sec^2 x - \tan^2 x = 1.
- f(x) = 1 for x > 0 and -1 for x < 0, so f'(x) = 0 at every x \ne 0. The domain is two intervals, and the zero-derivative result applies on each separately, giving a different constant on each.
- f'(x) = 3x^2 + 1 > 0, so f is increasing and takes each value at most once. f(1) = -3 < 0 < 5 = f(2), so by the intermediate value theorem there is a solution, between 1 and 2.
- Let \displaystyle h(x) = 1 + \frac{x}{2} - \sqrt{1 + x}, with h(0) = 0 and \displaystyle h'(x) = \frac{1}{2} - \frac{1}{2\sqrt{1 + x}} > 0 for x > 0. On [0, x], h(x) - h(0) = h'(c)\,x > 0.
- \lvert f(5) - f(1) \rvert \le 3 \cdot 4 = 12, so -8 \le f(5) \le 16. Both are possible: 4 - 3(x - 1) and 4 + 3(x - 1) reach them.
- \displaystyle \sqrt{101} - 10 = \frac{1}{2\sqrt{c}} for some c in (100, 101). Since \displaystyle 10 < \sqrt{c} < 11, the right side lies between \displaystyle \frac{1}{22} and \displaystyle \frac{1}{20}.
- (a) Five minutes is \displaystyle \frac{1}{12} hour, so his average speed is \displaystyle \frac{4}{1/12} = 48 miles per hour, and the mean value theorem gives an instant at that speed. (b) 4 \le 60(t_B - t_A), so \displaystyle t_B - t_A \ge \frac{1}{15} hour, 4 minutes: 3:04 p.m.
- The secant slope is \displaystyle \frac{p(b^2 - a^2) + q(b - a)}{b - a} = p(a + b) + q. Setting f'(c) = 2pc + q equal to it gives \displaystyle c = \frac{a + b}{2}.
- If f(x) = x^3 + 4x - 1 had two zeros r < s, Rolle’s theorem on [r, s] would give f'(c) = 3c^2 + 4 = 0, impossible since 3c^2 + 4 \ge 4. And f(0) = -1 < 0 < 4 = f(1), so there is a zero in (0, 1).
- Rolle’s theorem on [0, 1] and on [1, 2] gives c_1 in (0, 1) and c_2 in (1, 2) with f'(c_1) = f'(c_2) = 0. Since f' is differentiable, Rolle’s theorem for f' on [c_1, c_2] gives c with f''(c) = 0.
- Let d(t) be Biljana’s distance minus Goce’s. It is differentiable, 0 at the start and 0 at the finish, so Rolle’s theorem gives an instant with d'(t) = 0: equal speeds.
- For a \ne b the mean value theorem gives \sin b - \sin a = \cos c \,(b - a), and \lvert \cos c \rvert \le 1. For a = b both sides are 0.