# Section 4.2 Mean Value Theorem

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### Transcript of Section 4.2 Mean Value Theorem

Section 4.2 Mean Value TheoremWhat youll 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 UpGroup 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 TheoremShow 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

Example 1 continuedGiven f(x) = x2 Interval [0,2]

To use the MVTFind 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

Interpret your answerThe slope of the tangent line to f(x) = x2 at x = 1 is equal to the slope of the chord AB. ORThe 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 justmathtutoringMean Value Theorem

Warm UpPage 195Exercise 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.)

Show that the function satisfies the hypothesis of the MVT on the interval [0,1]. Then find cIs it continuous? Closed interval? Differentiable?

Find c

Interpret your findings

Example 2: Further Exploration of the MVTExplain 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 TheoremFind 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 cEvaluate 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 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

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?

HomeworkPage 202Exercises 1-11 Odds, 12-14

Todays AgendaPresent Homework on boardPage 202 Exercises 1-11 Odds, 12-14

4.2 Power PointExamples 5-8

Start todays homeworkPage 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).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 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 fallUse 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 fallWhere 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 3Find 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: AntiderivativeA 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 ruleAdd 1 to the exponentDivide the coefficient by (exponent + 1)Add CYou 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 dont forget C!f(x) = -sin x

f(x) = 2x + 6

f(x) = 2x2 + 4x - 3

Specific AntiderivativesFind the antiderivative ofthrough the point (0, 1).

Write equation for antiderivative + cInsert point (0,1) for (x, y) and solve for cWrite specific antiderivative.

Example 8 Finding Velocity and PositionWe 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 backwardsVelocity 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 backwardsVelocity 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.)

Its 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