Rolle’s theorem and Mean Value Theorem
(Section 3.2)
Alex Karassev
Rolle’s Theorem
y
y
f ′ (c) = 0
y = f(x)
y = f(x)
f(a) = f(b)
x
a
Example 1
c
b
x
a
c1
Example 2
c2
c3 b
Rolles Theorem

Suppose f is a function
such that
y
f ′ (c) = 0
f is continuous on [a,b]
 differentiable at least in (a,b)


If f(a) = f(b) then there
exists at least one c in (a,b)
such that f′(c) = 0
y = f(x)
x
a
c
b
Applications of Rolle’s Theorem

It is used in the prove of the
Mean Value Theorem

Together with the Intermediate Value
Theorem, it helps to determine exact number
of roots of an equation
Example

Prove that the equation cos x = 2x has
exactly one solution
y = 2x
y = cos x
Solution

cos x = 2x has exactly one solution means
two things:

It has a solution – can be proved using the IVT

It has no more than one solution – can be proved
using Rolle's theorem
cos x = 2x has a solution:
Proof using the IVT

cos x = 2x is equivalent to cos x – 2x = 0

Let f(x) = cos x – 2x

f is continuous for all x, so the IVT can be used

f(0) = cos 0 – 2∙0 = 1 > 0

f(π/2) = cos π/2 – 2 ∙ π/2 = 0 – π = – π < 0

Thus by the IVT there exists c in [0, π/2] such that
f(c) = 0, so the equation has a solution
f(x) = cos x – 2x = 0 has no more than one solution:
Proof by contradiction using Rolle's theorem

Assume the opposite: the equation has at least two solutions

Then there exist two numbers a and b s.t. a ≠ b and f(a) = f(b) = 0

In particular, f(a) = f(b)

f(x) is differentiable for all x, and hence Rolle's theorem is
applicable

By Rolle's theorem, there exists c in [a,b] such that f′(c) = 0

Find the derivative: f ′ (x) = (cos x – 2x) ′ = – sin x – 2

So if we had f ′ (c) = 0 it would mean that –sin c – 2 = 0 or
sin c = – 2, which is impossible!

Thus we obtained a contradiction

All our steps were logically correct so the fact that we obtained a
contradiction means that our original assumption "the equation
has at least two solutions" was wrong

Thus, the equation has no more than one solution!
f(x) = cos x – 2x = 0 has no more than one
solution: visualization
If it had two roots, then there would exist a ≠ b such that
f(a) = f(b) = 0
f ′ (c) = 0
y = f(x)
a
c
b
But f ′ (x) = (cos x – 2x) ′ = – sin x – 2
So if we had f ′ (c) = 0 it would mean that –sin c – 2 = 0 or
sin c = – 2, which is impossible
Mean Value Theorem
y
y
y = f(x)
y = f(x)
x
x
a c
Example 1
b
b
a
Example 2
There exists at least one point on the graph
at which tangent line is parallel to the secant line
Mean Value Theorem
f′(c) =
f(b) – f(a)
(b-a)

y
y = f(x)
f (b )  f ( a )
ba
x
a c
Slope of secant line is the
slope of line through the
points (a,f(a)) and (b,f(b)),
so it is
b

Slope of tangent line is f′(c)
MVT: exact statement
f′(c) =
f(b) – f(a)
(b-a)

Suppose f is continuous on
[a,b] and differentiable on (a,b)

Then there exists at least one
point c in (a,b) such that
y
y = f(x)
x
a c
b
f (b)  f (a)
f (c) 
ba
MVT: alternative formulations
f (b)  f (a)
f (c) 
ba

f (b)  f (a )  f (c)(b  a )

f (b)  f (a)  f (c)(b  a)
Interpretation of the MVT
using rate of change

Average rate of change
f (b )  f ( a )
ba
is equal to the instantaneous rate of change
f′(c) at some moment c

Example: suppose the cities A and B are connected by a straight
road and the distance between them is 360 km. You departed
from A at 1pm and arrived to B at 5:30pm. Then MVT implies that
at some moment your velocity v(t) = s′(t) was:
(s(5.5) – s(1)) / (5.5 – 1) = 360 / 4.5 = 80 km / h
Application of MVT
Estimation of functions
 Connection between the sign of derivative
and behavior of the function:
if f ′ > 0 function is increasing
if f ′ < 0 function is decreasing
 Error bounds for Taylor polynomials

Example


Suppose that f is differentiable for all x
If f (5) = 10 and f ′ (x) ≤ 3 for all x, how small
can f(-1) be?
Solution

MVT: f(b) – f(a) = f ′(c) (b – a) for some c in
(a,b)

Applying MVT to the interval [ –1, 5], we get:

f(5) – f(–1) = f ′(c) (5 – (– 1)) = 6 f ′(c) ≤ 6∙3 = 18

Thus f(5) – f(-1) ≤ 18

Therefore f(-1) ≥ f(5) – 18 = 10 – 18 = – 8