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Rolle’s theorem and Mean Value Theorem ( Section 4.2). Alex Karassev. Rolle’s Theorem. y. y. f ′ (c) = 0. y = f(x). y = f(x). f(a) = f(b). x. x. c 3. b. c 2. a. c 1. a. b. c. Example 1. Example 2. Rolles Theorem. Suppose f is a function such that f is continuous on [a,b]

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rolle s theorem
Rolle’s Theorem

y

y

f ′ (c) = 0

y = f(x)

y = f(x)

f(a) = f(b)

x

x

c3

b

c2

a

c1

a

b

c

Example 1

Example 2

rolles theorem
Rolles Theorem
  • Suppose f is a function such that
    • 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 ′ (c) = 0

y = f(x)

x

a

b

c

applications of rolle s theorem
Applications of Rolle’s Theorem
  • It is used in the prove of theMean Value Theorem
  • Together with the Intermediate Value Theorem, it helps to determine exact number of roots of an equation
example

y = cos x

Example
  • Prove that the equation cos x = 2x has exactly one solution

y = 2x

solution
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
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 thatf(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

f(x) = cos x – 2x = 0 has no more than one solution:
  • Assume the opposite: the equation has at least two solutions
  • Then there exist two numbers a and b s.t. a ≠ b andf(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 orsin 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
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 thatf(a) = f(b) = 0

f ′ (c) = 0

y = f(x)

b

a

c

But f ′ (x) = (cos x – 2x) ′ = – sin x – 2

So if we had f ′ (c) = 0 it would mean that –sin c – 2 = 0 orsin c = – 2, which is impossible

mean value theorem

y

y

y = f(x)

y = f(x)

x

x

b

a

b

a

c

Example 2

Example 1

Mean Value Theorem

There exists at least one point on the graphat which tangent line is parallel to the secant line

mean value theorem1

f(b) – f(a) (b-a)

f′(c) =

Mean Value Theorem
  • Slope of secant line is the slope of line through the points (a,f(a)) and (b,f(b)), so it is
  • Slope of tangent line is f′(c)

y

y = f(x)

x

b

a

c

mvt exact statement

f(b) – f(a) (b-a)

f′(c) =

MVT: exact statement
  • 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

b

a

c

interpretation of the mvt using rate of change
Interpretation of the MVTusing rate of change
  • Average rate of changeis equal to the instantaneous rate of changef′(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
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
example1
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?
solution1
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