Rolles Theorem
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Transcript Rolles Theorem
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 )
ba
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)
ba
MVT: alternative formulations
f (b) f (a)
f (c)
ba
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 )
ba
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