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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] - PowerPoint PPT Presentation
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Rolle’s theorem and Mean Value Theorem(Section 4.2)
Alex Karassev
Rolle’s Theorem
x
y
y = f(x)
ca bx
y
y = f(x)
c1 c2a b
f(a) = f(b)
c3
f ′ (c) = 0
Example 1 Example 2
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
x
y
y = f(x)
ca b
f ′ (c) = 0
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
y = cos x
Example
Prove that the equation cos x = 2x has exactly one solution
y = 2x
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 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
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
Proof using the IVT
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 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!
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!
Proof by contradiction using Rolle's theorem
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
y = f(x)
ca b
f ′ (c) = 0
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
x
y
y = f(x)
a b
Example 2
x
y
y = f(x)
ca b
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 Theorem
x
y
y = f(x)
ca b
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)
f(b) – f(a)(b-a)
f′(c) =
ab
afbf
)()(
MVT: exact statement
x
y
y = f(x)
ca b
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
f(b) – f(a)(b-a)
f′(c) =
ab
afbfcf
)()()(
MVT: alternative formulations
))(()()(
))(()()(
)()()(
abcfafbf
abcfafbf
ab
afbfcf
Interpretation of the MVTusing rate of change
Average rate of change
is 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
ab
afbf
)()(
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
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