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The Mean Value Theorem
Astudent at the University of Connecticut happens to be travelling
to Boston. He enters the Massachussetts Turnpike at the Sturbridge
Village entrance at 9:15 in the morning. Since he uses Fast Lane, he
never actually picks up a ticket, but sensors record that he goes through
the toll lane at Newton at 9:53, just 38 minutes later.
He soon receives a traffic summons in the mail indicating that he vi-
olated the posted speed limit and decides to appeal, since there is no
indication he was caught by radar.
Appearingincourt, theprosecutorarguesthathetravelled44.8miles
in 38 minutes and therefore travelled at an average speed of 70.74 miles
per hour, more than 5 miles above the highest posted speed limit of 65
miles per hour.
Thestudent argues that, although he may have averaged more than 65
miles per hour, there is no evidence that he actually ever travelled at
an instantaneous speed of more than 65 miles per hour.
The judge considers the arguments and quickly rejects the student’s
plea, noting that the Mean Value Theorem implies that since his aver-
age speed was 70.74 miles per hour, there had to be at least one instant
during which his instantaneous speed was 70.74 miles per hour.
Rolle’s Theorem
Theorem1(Rolle’sTheorem). Suppose a function f is continuous on
the closed interval [a,b], differentiable on the open interval (a,b) and
f(a) = f(b). Then there is a number c ∈ (a,b) such that f′(c) = 0.
Geometrically, this says that if there are two points on a smooth curve
takes at the same height, there must be a point in between where the
tangent is horizontal.
Rolle’s Theorem is a special case of the Mean Value Theorem.
Proof: Suppose f satisfies the hypotheses of Rolle’s Theorem. By the
Extreme Value Theorem for Continuous Function, there must be some
point in [a,b] at which f attains a minimum and some point at which
f attains a maximum.
One possibility is that f is constant on the entire interval, in which
case f′ is identically 0 on (a,b) and the conclusion is clearly true.
So let’s consider the other possibility, that f is not constant. Then
either the minimum or the maximum must occur at some point c ∈
(a,b), that is, at some point other than the endpoints.
1
2
Wewillshowthatf′(c) = 0 if f has a maximum at c. Similar reasoning
would show f′(c) = 0 if f had a minimum at c, showing the conclusion
is true and completing the proof.
Weknow f′(c) = lim f(x)−f(c).
x→c x−c
Since this ordinary limit exists, it follows that both the left hand limit
and the right hand limit exist and are both equal to f′(c). We’ll con-
sider them separately. f(x)−f(c)
The Left Hand Limit: f′(c) = lim − .
x→c x−c
Sincef hasamaximumatc,ifx < c,thenf(x) ≤ f(c),sof(x)−f(c) ≤
0. But, if x < c, it’s also true that x − c < 0 and it follows that
f(x)−f(c) ≥ 0. We see lim − f(x)−f(c) is the limit of non-
x−c x→c x−c
negative numbers and therefore can’t be negative. It follows that
f′(c) ≥ 0.
The line of reasoning for the right hand limit is similar.
The Right Hand Limit: f′(c) = lim + f(x)−f(c).
x→c x−c
Sincef hasamaximumatc,ifx > c,thenf(x) ≤ f(c),sof(x)−f(c) ≤
0. But, if x > c, it’s also true that x − c > 0 and it follows that
f(x)−f(c) ≤ 0. We see lim + f(x)−f(c) is the limit of numbers
x−c x→c x−c
less than or equal to 0 and therefore can’t be positive. It follows that
f′(c) ≤ 0.
Since f′(c) ≥ 0 and f′(c) ≤ 0, it follows that f′(c) = 0 QED
Consequences of Rolle’s Theorem
Besides being a special case of the Mean Value Theorem and being a
step in the path to proving the Mean Value Theorem, Rolle’s Theorem
has some interesting applications of its own.
Corollary 2. A polynomial equation of degree n has at most n solu-
tions.
This is equivalent to the statement that a polynomial of degree n has
at most n zeros.
Since a polynomial may be differentiated as many times as necessary,
with each derivative being a polynomial of lower degree, one immediate
consequence of Rolle’s Theorem is that the derivative of a polynomial
has at least one zero between each pair of distinct zeros of the original
polynomial.
3
The derivative of a linear polynomial is a non-zero constant, having no
zeros, so a linear polynomial can’t have more than 1 zero.
Thederivative of a quadratic polynomial is linear, having no more than
1 zero, so a quadratic can’t have more than 2 zeros.
Thederivative of a cubic polynomial is quadratic, having no more than
2 zeros, so the cubic can’t have more than 3 zeros.
This clearly goes on forever. The argument can be made rigorous
through the use of Mathematical Induction.
The Mean Value Theorem
Theorem 3 (The Mean Value Theorem). Suppose a function f is
continuous on the closed interval [a,b] and differentiable on the open
interval (a,b). Then there is a number c ∈ (a,b) such that f(b)−f(a) =
f′(c)(b − a) or, equivalently, f′(c) = f(b) − f(a).
b −a
Geometrically, the Mean Value Theorem says that if there is a smooth
curve connecting two points, there must be some point in between
at which the tangent line is parallel to the line connecting those two
points.
Analytically, the Mean Value Theorem says the rate of change of a
differentiable function must, at some point, take on its average, or
mean value.
The proof of the Mean Value Theorem depends on the fact that
the particular point at which the tangent line is parallel to the line
connecting the endpoints also happens to be the point at which the
curve is furthers away from that line. The proof essentially consists
of applying Rolle’s Theorem to the function measuring the distance
between the line and the curve.
Since the line goes between the points (a,f(a)) and (b,f(b)), its
slope will be f(b) − f(a) and its equation may be written, in point-
slope form, b −a
y −f(a) = f(b)−f(a) ·(x−a).
b −a
Solving for y, we may write the equation in the form
y = f(a)+ f(b)−f(a) ·(x−a).
b −a
Since the second coordinate of a point on the curve with first coordinate
xis f(x), the vertical distance between the line and the curve will equal
4
f(b) −f(a)
f(x)− f(a)+ b −a · (x − a) =
f(x)−f(a)− f(b)−f(a) ·(x−a).
b −a
Weare now prepared to prove the Mean Value Theorem.
Proof: Let φ(x) = f(x)−f(a)− f(b)−f(a) ·(x−a).
b −a
It is easy to see that φ satisfies the hypotheses of Rolle’s Theorem on
the interval [a,b]. Certainly, the fact that φ is both continuous and
differentiable on [a,b] follows immediately from the fact that f is. In
addition,
φ(a) = f(a)−f(a)− f(b)−f(a) ·(a−a) = 0
and b −a
φ(b) = f(b)−f(a)−f(b)−f(a)·(b−a) = f(b)−f(a)−[f(b)−f(a)] = 0.
b −a ′
Thus,′there must be some c ∈ (a,b) such that φ (c) = 0. We first
obtain φ (c) as follows.
′ ′ f(b) −f(a)
φ(x) = f (x)− b −a
′ ′ f(b) −f(a)
φ(c) = f (c) − b −a
′
Since φ (c) = 0, it follows that
f′(c) − f(b) − f(a) = 0.
b −a
f′(c) = f(b) − f(a)
′ b −a
f (c)(b − a) = f(b) − f(a) QED
Consequences of the Mean Value Theorem
PerhapsthemostimportantconsequenceoftheMeanValueTheorem
is that it gives precise meaning to the most important single concept
in elementary Calculus,
The Derivative Measures Rate of Change.
Theorem 4.
a. If the derivative of a function is positive at all points on an
interval, then the function is increasing on that interval.
b. If the derivative of a function is negative at all points on an
interval, then the function is decreasing on that interval.
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