Transcript Lesson 6-2 - Headlee's Math Mansion

Lesson 4-2
Mean Value Theorem and
Rolle’s Theorem
Quiz
• Homework Problem: Related Rates 3-10
Gravel is being dumped from a conveyor belt at
a rate of 30 ft³/min, and forms a pile in shape of
a cone whose base diameter and height are
always equal. How fast is the height of the pile
increasing when the pile is 10ft high?
• Reading questions:
– What were the names of the two theorems in section
4.2?
– What pre-conditions (hypotheses) do the Theorems
have in common?
Objectives
• Understand Rolle’s Theorem
• Understand Mean Value Theorem
Vocabulary
• Existence Theorem – a theorem that guarantees that
there exists a number with a certain property, but it
doesn’t tell us how to find it.
Theorems
Mean Value Theorem: Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
then there is a number c in (a,b) such that
f(b) – f(a)
f’(c) = --------------- or equivalently: f(b) – f(a) = f’(c)(b – a)
b–a
Rolle’s Theorem: Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
c) f(a) = f(b)
then there is a number c in (a,b) such that f’(c) = 0
Mean Value Theorem (MVT)
Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
then there is a number c in (a,b) such that
f(b) – f(a)
f’(c) = --------------b–a
(instantaneous rate of change, mtangent = average rate of change, msecant)
f(b) – f(a) = f’(c)(b – a)
or equivalently:
P(c,f(c))
y
y
P1
A(a,f(a))
P2
B(b,f(b))
a
c
b
x
a
c1
c2
b
For a differentiable function f(x), the slope of the secant line through (a, f(a)) and (b,
f(b)) equals the slope of the tangent line at some point c between a and b. In other
words, the average rate of change of f(x) over [a,b] equals the instantaneous rate of
change at some point c in (a,b).
x
Example 1
Verify that the mean value theorem (MVT) holds for
f(x) = -x² + 6x – 6 on [1,3].
a) continuous on the closed interval [a,b]  polynomial
b) differentiable on the open interval (a,b)  polynomial
f’(x) = -2x + 6
f(b) – f(a) = f(3) – f(1) = 3 – (-1) = 4
f(b) – f(a) / (b – a) = 4/2 = 2
f’(x) = 2 = 6 – 2x
so -4 = -2x
2=x
Example 2
Find the number that satisfies the MVT on the given
interval or state why the theorem does not apply.
f(x) = x2/5 on [0,32]
a) continuous on the closed interval [a,b]  ok
b) differentiable on the open interval (a,b)  ok on open
f’(x) = (2/5)x-3/5
f(b) – f(a) = f(32) – f(0) = 4 – 0 = 4
f(b) – f(a) / (b – a) = 4/32 = 0.125
f’(x) = 0.125 = (2/5)x-3/5 so x = 6.94891
Example 3
Find the number that satisfies the MVT on the given
interval or state why the theorem does not apply.
f(x) = x + (1/x) on [1,3]
a) continuous on the closed interval [a,b]  ok
b) differentiable on the open interval (a,b)  ok on open
f’(x) = 1 – x-2
f(b) – f(a) = f(3) – f(1) = 10/3 – 2 = 4/3
f(b) – f(a) / (b – a) = (7/3)/2 = 2/3
f’(x) = 2/3 = 1 – x-2
so x-2 = 1/3 x² = 3
x=3 = 1.732
Example 4
Find the number that satisfies the MVT on the given
interval or state why the theorem does not apply.
f(x) = x1/2 + 2(x – 3)1/3 on [1,9]
a) continuous on the closed interval [a,b]  ok
b) differentiable on the open interval (a,b)  vertical tan
f’(x) = 1/2x-1/2 + 2/3(x – 3)-2/3
f’(x) undefined at x = 3 (vertical tangent)
MVT does not apply
Rolle’s Theorem
Let f be a function that is
a) continuous on the closed interval [a,b]
b) differentiable on the open interval (a,b)
c) f(a) = f(b)
then there is a number c in (a,b) such that f’(c) = 0
y
y
A.
y
C.
B.
a
c
bx
y
a
c
bx
D.
a c1
c2
bx
a
c
bx
Case 1: f(x) = k (constant) [picture A]
Case 2: f(x) > f(a) for some x in (a,b) [picture B or C]
Extreme value theorem guarantees a max value somewhere in [a,b]. Since
f(a) = f(b), then at some c in (a,b) this max must occur. Fermat’s Theorem s
that f’(c) =0.
Case 3: f(x) < f(a) for some x in (a,b) [picture C or D]
Extreme value theorem guarantees a min value somewhere in [a,b]. Since
f(a) = f(b), then at some c in (a,b) this min must occur. Fermat’s Theorem s
that f’(c) =0.
Example 5
Determine whether Rolle’s Theorem’s hypotheses are
satisfied &, if so, find a number c for which f’(c) = 0.
f(x) = x² + 9 on [-3,3]
a) continuous on the closed interval [a,b]  polynomial
b) differentiable on the open interval (a,b)  polynomial
c) f(a) = f(b)
 f(-3) = 18 f(3) = 18
f’(x) = 2x
f’(x) = 0 when x = 0
0 in the interval [-3,3]
Example 6
Determine whether Rolle’s Theorem’s hypotheses are
satisfied &, if so, find a number c for which f’(c) = 0.
f(x) = x³ - 2x² - x + 2 on [-1,2]
a) continuous on the closed interval [a,b]  polynomial
b) differentiable on the open interval (a,b)  polynomial
c) f(a) = f(b)
 f(-1) = 0 f(2) = 0
f’(x) = 3x² - 4x – 1
f’(x) = 0 when x = (7 + 2)/3 = 1.5486
f’(x) = 0 when x = -(7 - 2)/3 = -0.2153
-0.2153 and 1.5486 in the interval [-1,2]
Example 7
Determine whether Rolle’s Theorem’s hypotheses are
satisfied &, if so, find a number c for which f’(c) = 0.
f(x) = (x² - 1) / x on [-1,1]
a) continuous on the closed interval [a,b]  no at x = 0
b) differentiable on the open interval (a,b)  no at x = 0
c) f(a) = f(b)
 f(-1) = 0 f(1) = 0
Example 8
Determine whether Rolle’s Theorem’s hypotheses are
satisfied &, if so, find a number c for which f’(c) = 0.
f(x) = sin x on [0,π]
a) continuous on the closed interval [a,b]  ok
b) differentiable on the open interval (a,b)  ok
c) f(a) = f(b)
 f(0) = 0 f(π) = 0
f’(x) = cos x
f’(x) = 0 (or undefined) when x = π/2
π/2 in the interval [0,π]
Summary & Homework
• Summary:
– Mean Value and Rolle’s theorems are
existance theorems
– Each has some preconditions that must be
met to be used
• Homework:
– pg 295-296: 2, 7, 11, 12, 24