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Section 4.2 Mean Value Theorem
What you’ll learnMean Value TheoremPhysical InterpretationIncreasing and Decreasing FunctionsOther Consequences
Why?The Mean Value Theorem is an important theoretical tool to connect the average and instantaneous rate of change.
Warm Up
Group Activity:
AP Test Prep Series Pages 131 – 133
Due in 15 minutes
Correct Homework
This means…….
IFy = f(x) is continuousy = f(x) is on a closed interval [a,b]y = f(x) is differentiable at every point in its interior
(a,b)
THENSomewhere between points A and B on a
differentiable curve, there is at least one tangent line parallel to chord AB.
http://justmathtutoring.com/ Mean Value Thm.
Example 1: Exploring the Mean Value Theorem
Show that the function f(x) = x2 satisfies the hypothesis of
the Mean Value Theorem on the interval [0,2]. Then find a solution c to the equation
Consider f(x) = x2. Is it continuous? Closed interval? Differentiable?
If so, by the MVT we are guaranteed a point c in the interval [0,2] for which ab
afbfcf
)()(
)('
ab
afbfcf
)()(
)('
Example 1 continued
Given f(x) = x2
Interval [0,2]
To use the MVT Find the slope of the chord
with endpoints (0, f(0)) and (2, f(2)).
Find f’
Set f’ equal to the slope of the chord, solve for c
Find c
ab
afbfcf
)()(
)('
Interpret your answer
The slope of the tangent line to f(x) = x2 at
x = 1 is equal to the slope of the chord AB.
OR The tangent line at x = 1 is parallel to chord AB.
Write equations for line AB and the tangent line of y = x2 at x=1.
Graph and investigate. The lines should be parallel.
Homework
Lesson 4.2
Quick Review Ex 1-10
Watch justmathtutoring
Mean Value Theorem
Warm Up
Page 195
Exercise 51 a-c
Theorem 3: Mean Value Theorem for Derivatives If y = f(x) is continuous at every point on the
closed interval [a, b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which
(The derivative at some point c = the slope of the chord.)
ab
afbfcf
)()(
)('
Show that the function satisfies the hypothesis of the MVT on the interval [0,1]. Then find c
1. Is it continuous? Closed interval? Differentiable?
2. Find c
3. Interpret your findings
12)( 2 xxxf
Example 2: Further Exploration of the MVT
Explain why each of the following functions fails to satisfy the conditions of the Mean Value Theorem on the interval [-1,1].
You try:
1)( 2 xxf1,1
1,3{)(
2
3
xx
xxxf
3
1
)( xxf
Example 3: Applying the Mean Value Theorem
Find a tangent to f in the interval (-1,1) that is parallel to the secant AB.
Given: , A = (-1,f(-1)) and B = (1, f(1))
1) Find the slope of AB and f’(c)
2) Apply MVT to find c
3) Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line
21)( xxf
You Try -
1) Find slope of chord AB that connects endpoints
2) Find f ’ and apply MVT formula to find c.
3) Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line. Graph & check.
31 x1)( xxf
Physical Interpretation of the Mean Value TheoremThe MVT says the instantaneous change at
some interior point must equal the average change over the entire interval.
f’(x) = instantaneous change at a point
= average change over the
interval
ab
afbf
)()(
Example 4: Interpreting the MVTIf a car accelerating from zero takes 8 sec to go
352 ft, its average velocity for the 8-second interval is 352 / 8 = 44 ft/sec, or 30 mph.
Can we cite the driver for speeding if he / she is in a residential area with a speed limit of 25 mph?
hour
miles
hourft
mileft 30
1
sec3600
5280
1
sec
44
Homework
Page 202
Exercises 1-11 Odds, 12-14
Today’s Agenda
Present Homework on board
Page 202 Exercises 1-11 Odds, 12-14
4.2 Power Point
Examples 5-8
Start today’s homework
Page 203 Exercises 15-33 odds, 37, 43, 45
4.2 Continued:Increasing Functions / Decreasing Functions
Let f be a function defined on an interval I and
let x1 and x2 be any two points on I
f increases on I if x1 < x2 => f(x1) < f(x2)
f decreases on I if x1 < x2 => f(x1) > f(x2)
Corollaries to the Mean Value TheoremCorollary 1: Increasing & Decreasing Functions
Let f be continuous on [a,b] and differentiable on (a,b).1) If f ’ > 0 at each point of (a,b), then f increases on [a,b]2) If f ’< 0 at each point of (a,b), then f decreases on [a,b]
Corollary 2: Functions with f’ = 0 are ConstantIf f ’(x) = 0 at each point of an interval I, then there is a constant C
for which f(x) = C for all x in I
Corollary 3: Functions with the same derivative differ by a constant.If f ’(x) = g ’(x) at each point of an interval I, then there is a constant
C such that f(x) = g(x) + C for all x in I.
http://justmathtutoring.com/ Using 1st derivative to find where f is increasing or decreasing, find max or min points
Example 5 Determining where graphs rise or fall
Use corollary 1 to determine where the graph of
f(x) = x2 – 3x is increasing and decreasing.
1) Find f’ and set it equal to zero to find critical points.
2) Where f ’ > 0, f is increasing.3) Where f ’ < 0, f is decreasing.4) Any maximum or minimum values?
Example 6 Determining where graphs rise or fall
Where is the function increasing and where is it decreasing?
Graphically: Use window [-5,5] by [-5,5]Confirm Analytically: Find f ’, evaluate f ’ = 0 to find critical points.
Where f ’ > 0, f is increasing. Where f ’ < 0, f is decreasing.
xxxf 4)( 3
Example 7 Applying Corollary 3
Find the function f(x) whose derivative is sin x and whose graph passes through the point (0,2).
Write f(x) = antiderivative function + C Use (x,y) = (0, 2) in equation, solve for C.Write f(x), the antiderivative of sin x through (0, 2)
Definition: Antiderivative
A function F(x) is an antiderivative of a function f(x) if F’(x) = f(x) for all x in the domain of f. The process of finding an antiderivative is antidifferentiation.
If you are given a derivative function, “Think Backwards” to get the original using all of the chapter 3 differentiation relationships.
Find the Antiderivative!
Find the antiderivative of
f(x) = 6x2.
Reverse the power rule Add 1 to the exponent Divide the coefficient by
(exponent + 1) Add C You have found the
antiderivative, F(x)
Find the antiderivative of
f(x) = cos x
Think Backwards: What function has a derivative of cos x?
Add C
You have found the antiderivative, F(x).
Find each antiderivative – don’t forget C!1) f’(x) = -sin x
2) f’(x) = 2x + 6
3) f’(x) = 2x2 + 4x - 3
Specific Antiderivatives
Find the antiderivative of
through the point (0, 1).
Write equation for antiderivative + c Insert point (0,1) for (x, y) and solve for c Write specific antiderivative.
3
1
3
2)( xxf
Example 8 Finding Velocity and Position
We can use antidifferentiation to find the velocity and position functions of a body falling freely from a height of 0 meters under each of the following sets of conditions.
a) The acceleration is 9.8 m/sec2 and the body falls from rest.
b) The acceleration is 9.8 m/sec2 and the body is propelled downward with an initial velocity of 1 m/sec2.
The acceleration is 9.8 m/sec2 and the body falls from rest.
Work backwards
Velocity function + c, P(0,0) (why?)
Position function + c (from velocity function)
The acceleration is 9.8 m/sec2 and the body is propelled downward with an initial velocity of 1 m/sec2.
Work backwards
Velocity function + c, P(0,1) (why?)
Position function + c (from velocity function)
SummaryThe Mean Value Theorem tells us that If y = f(x) is continuous at every point on the closed interval [a, b] and differentiable at
every point of its interior (a,b), then there is at least one point c in (a,b) at which(The derivative at point c = the slope of the chord.)
It’s corollaries go on to tell us that where f ‘ is positive, f is increasing, where f ‘ = 0, f is a constant function, and where f ‘ is negative, f is decreasing.
We can work backwards from a derivative function to the original function, a process called antidifferentiation. However, as the derivative of any constant = 0, we need to know a point of the original function to get its specific antiderivative. Without a point of the function all we can determine is a general formula for f(x) + C.
Homework
Page 203Exercises 15-33 odds,
37, 43, 45
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