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Section 4.2 Mean Value TheoremPowerPoint Presentation

Section 4.2 Mean Value Theorem

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### Section 4.2 Mean Value Theorem

What you’ll learn

Mean Value Theorem

Physical Interpretation

Increasing and Decreasing Functions

Other Consequences

Why?

The Mean Value Theorem is an important theoretical tool to connect the average and instantaneous rate of change.

This means…….

IF

y = f(x) is continuous

y = f(x) is on a closed interval [a,b]

y = f(x) is differentiable at every point in its interior (a,b)

THEN

Somewhere 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.

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

Example 1: Exploring the Mean Value TheoremInterval [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

Example 1 continuedInterpret 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.

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.)

Show that the function satisfies the hypothesis of the MVT on the interval [0,1]. Then find c

- Is it continuous? Closed interval? Differentiable?
- Find c
- Interpret your findings

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:

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))

- Find the slope of AB and f’(c)
- Apply MVT to find c
- Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line

You Try -

- Find slope of chord AB that connects endpoints
- Find f ’ and apply MVT formula to find c.
- Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line. Graph & check.

Physical Interpretation of the Mean Value Theorem

The 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

Example 4: Interpreting the MVT

If 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?

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 Theorem

Corollary 1: Increasing & Decreasing FunctionsLet f be continuous on [a,b] and differentiable on (a,b).

- If f ’ > 0 at each point of (a,b), then f increases on [a,b]
- If f ’< 0 at each point of (a,b), then f decreases on [a,b]
Corollary 2: Functions with f’ = 0 are Constant

If 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.

- Find f’ and set it equal to zero to find critical points.
- Where f ’ > 0, f is increasing.
- Where f ’ < 0, f is decreasing.
- 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.

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.

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 the Antiderivative!Find each antiderivative – don’t forget C!

- f’(x) = -sin x
- f’(x) = 2x + 6
- 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.

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.

- The acceleration is 9.8 m/sec2 and the body falls from rest.
- 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)

Summary

The 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.

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