Chapter 3 – Spring 2014

Chapter 3 – Spring 2014

Section 3.1 – Critical Numbers and Absolute Extrema on an Interval

Critical Numbers

Critical numbers are x values on a continuous function at which f’(x) = 0 or f’(x) DNE.

What would these places look like on a graph?

1. Horizontal Tangents (Min/Max/half and half)2. Vertical Tangents3. Cusp

Find the critical numbers of the graph graphically, then algebraically

Critical Numbers: x=0 and x=.816

2

3( ) 2 3f x x x

Uses of critical numbers

Critical numbers are locations where relative minimums and maximums are possible.

They are also possible locations for absolute extrema of a function.

Locating Absolute Extrema on a closed intervalThe location of the absolute minimum and absolute maximum of a function on a closed interval must be located at one of two places.

1.Critical Numbers2.Endpoints

Find the absolute extrema of on the interval [2,6]

2

27( )

2f x x

x

Critical Numbers and Absolute Extrema HomeworkP 169 (13,15,16,21,23,30)

Section 3.2 – Rolle’s Theorem and Mean Value

Theorem

Rolle’s Theorem

Rolle’s Theorem states that on a continuous and differentiable interval between b and c, If f(b) = f(c), Then there is at least one number in between (we’ll call it d) where f’(d) = 0.Stated otherwise, if you place two points on a graph with the same y value, no matter how you connect them there will be at least one point in between with a horizontal tangent.

Why is Rolle’s Theorem Helpful

Rolle’s Theorem is typically used to prove that a function has to have a relative minimum or a relative minimum. This is helpful when finding a derivative and setting it equal to zero is very cumbersome.

Using Rolle’s Theorem

Consider the polynomial f(x)=(x-1)(x-3)(x-6)

Using Rolle’s Theorem, explain why f(x) must have at least two locations where f’(x) = 0.

One what intervals are the horizontal tangents located? (___,___) and (___,___)

Places Where Rolle’s Theorem Does Not ApplyBelow is the graph of f(x) = cot(x) from –π to π . f(-π/2)=0 and f(π/2)=0. Rolle’s Theorem states that if you place two points on a graph with the same y value, no matter how you connect them there will be at least one point in between with a horizontal tangent.Why does Rolle’s Theorem not apply here?

The graph is not continuous.

Places Where Rolle’s Theorem Does Not ApplyThe graph below is f(x) = lx-2l+1. Why does Rolle’s Theorem not apply here?

The graph is not differentiable atall point on the interval.

Determine whether Rolle’s Theorem applies to the function. If it does, write any intervals on which it applies. Then find all values of c on your intervals at which f’(c)=0

32( ) 1f x x

Determine whether Rolle’s Theorem applies to the function. If it does, write any intervals on which it applies. Then find all values of c on your intervals at which f’(c)=0

2( ) ( 3)( 1)f x x x

Rolle’s Theorem Homework

Rolle’s Theorem: P 176 (1,2,8,11,18,19,29)

Mean Value Theorem

The mean value theorem is an extension of Rolle’s Theorem.

It states that on an interval (a,b) where the function is continuous and differentiable, the exists some point c where

Stated otherwise, if you connect the to endpoints of a curve and find the slope, there is at least one point on the curve who derivative matches that slope.

( ) ( )'( )

f b f af c

b a

Illustration of the Mean Value Theorem

According to the mean value theorem there must be at least one other point between a and b with a derivative equal to that slope.

Our eyes can easily confirm this is true by locating the tangent line with the same slope.

Using the Mean Value Theorem

Given the function, f(x) = 3 – 8/x, does the Mean Value Theorem apply on the interval (4,8). If it does, find the values of c where

The mean value theorem does apply on the interval (4,8). f’(c) = ¼ at c=5.66

( ) ( )'( )

f b f af c

b a

Using the Mean Value Theorem

Given the function, f(x) = , does the Mean Value Theorem apply on the interval (0,4). If it does, find the values of c where

The mean value theorem does not apply on the interval (0,4) because function is not continuous or differentiable at all points.

( ) ( )'( )

f b f af c

b a

2

2 1

x

x

Places where the Mean Value Theorem does not applyThe Mean Value Theorem has the same conditions as Rolle’s Theorem.

1.The function must be continuous.2.The function must be differentiable.

Using the Mean Value Theorem - SpeedingPolice recently installed two traffic cameras for recording license plate numbers on a highway where the speed limit is 65 MPH. You are the clerk for a county judge working on a case where a man received the following ticket in the mail. The man is contesting the ticket on the grounds that at no point did the police actually record him speeding with a radar gun. Your job is to provide a legal basis for the ticket to your boss.

Speeding Violation - $235 1/3/14License Plate BQG-923Mile Marker 13 @ 10:12:00 AMMile Marker 23 @ 10:20:30 AM

Mean Value Theorem HW

Mean Value Theorem: P 177 (37,40,42,43,58,59)

Section 3.3 – Increasing and Decreasing Functions and the First Derivative

Test

What is the Domain of the function2

2

2(x 9)( )

4f x

x

Domain can be written in two ways: 1. Based on what is defined 2. Based on what isn’t defined. In Calculus we usually write the domain based on what is defined.

Increasing and Decreasing FunctionsIf f(x) is a continuous and differentiable function on the interval (a,b),Then:

1.f’(x) > 0 for all x in (a,b) iff f(x) is increasing on the interval (a,b)2.f’(x) < 0 for all x in (a,b) iff f(x) is decreasing on the interval (a,b)3.f’(x) = 0 for all x in (a,b) iff f(x) is constant on the interval (a,b)

Where is the function increasing or deceasing?

2

2

2(x 9)( )

4f x

x

We describe where the function is increasing and decreasing the same way we write domain of a function

Determine where the function is increasing and decreasing without a graph.

Functions change between increasing and decreasing at critical numbers, so let’s find the critical numbers and then make a chart.

The critical numbers divide our intervals.

3 23( )

2f x x x

Interval

Test #

Sign of f’(Test #)

Conclusion

First Derivative TestThe first derivative test requires an understanding of increasing and decreasing functions to determine relative minimums and maximums.

A relative maximum is a point where the function is increasing on the left and decreasing on the right.

A relative minimum is a point where the function is decreasing on the left and increasing on the right.

Determine where the function is increasing and decreasing. Then apply the first derivative test to determine the location of any relative minimums or maximums.

The critical numbers divide the intervals.

13( )f x x x

Interval

Test #

Sign of f’(Test #)

Conclusion

Functions change between increasing and decreasing at critical numbers, so let’s find the critical numbers and then make a chart.

Curve Sketching Using the Derivative

x -3 -2 -1 0 1 2 3

f’(x) 10 3 0 -1 0 3 10

Remember f’(x) provides the slopes, not y values

Curve Sketching Using the Derivative

x -3 -2 -1 0 1 2 3

f’(x) -2 0 2 ∞ 2 0 -2

Increasing, Decreasing, and First Derivative Test Homework.Increasing and Decreasing Functions: P 186 (3,6,7,11,13.80)

First Derivative Test: P 186 (21,28,33,35,76)

Section 3.4 – Concavity and the Second Derivative

Concavity: Concave Up vs. Concave Down

Relationship between concavity and the second derivative

When a function is concave down the first derivative is decreasing, so the second derivative is negative.

When a function is concave up the first derivative is increasing, so the second derivative is positive.

Using Faces to Remember the Concavity and 2nd Derivative Connection

Describe the concavity of a function.

We express the concavity of a function the same way we express domain, increasing, and decreasing by using interval notation.

Concave Up:

Concave Down:

Inflection Point

On a continuous function, a point is an inflection point if the graph changes from concave up to concave down (or vise versa) at that location.

An inflection point can also be described at a location where to second derivative changes from positive to negative (or vise versa).

At an inflection point, f’’(x)=0 or it does not exist.

Locate the Inflection Points on the GraphInflection Points:

Find the Inflection Points of

Let’s find the second derivative because either f’’(x)=0 or does not exist at an inflection point.

4 3( ) 4f x x x

Inflection points are x=0 and x=2

Determine the intervals on which the function below is concave up and concave down.

Concavity changes when f’’(x)=0 or f’’(x) DNE. Let’s find these places and then make a chart.

3 2( ) 2 3 12 5f x x x x

Interval

Test #

Sign of f’’(Test #)

Conclusion

Determine the intervals on which the function below is concave up and concave down.Concavity changes when f’’(x)=0 or f’’(x) DNE. Let’s find

these places and then we’ll make a chart on the next slide.2

2

1( )

4

xf x

x

There are no points where f’’(x)=0, but f’’ DNE at x=-2 and x=2

Interval

Test #

Sign of f’’(Test #)

Conclusion

2

2

1( )

4

xf x

x

Second Derivative Test

The second derivative test applies the following understanding to determine which critical numbers are relative minimums and which are relative maximums.

A relative maximum is always located on a section of the graph that is concave down, f’’(x) < 0

A relative minimum is always located on a section of a graph that is concave up, f’’(x) > 0

Find all Relative Extrema. Apply the Second Derivative Test to determine Max’s from Min’s.

5 3( ) 3 5f x x x Relative extrema are located at critical numbers. Let’s locate them, then apply the

second derivative test to see which are relative maximums or minimums.

Applying the 2nd Derivative Test

Point

Sign of f”(point)

Conclusion

5 3( ) 3 5f x x x

Sketch a Graph with the following Characteristics• f(1) = f(5) = 0• f’(x) < 0 when x < 3• f’(3) = 0• f’’(x) < 0 for all values of x.

Sketch a Graph with the following Characteristics• f’(x) > 0 if x < 2• f’(x) < 0 if x > 2• f’’(x) < 0 while x ≠2• f(2) is undefined & f’(2) DNE

Sketch a Graph with the following Characteristics• f(1) = 2• f’’(1) = 0• f’(x) < 0 for all values of x• f’’< 0 if x < 1

Find a, b, c, & d so the cubic satisfies the given conditions.Relative Maximum: (-1,1)Relative Minimum: (1,-1)Inflection Point: (0,0)

3 2( )f x ax bx cx d

Find a, b, c, & d so the cubic satisfies the given conditions.Relative Maximum: (2,2)Relative Minimum: (0,-2)Inflection Point: (1,0)

3 2( )f x ax bx cx d

Concavity, Inflection Point, and 2nd Derivative Test HomeworkConcavity, Inflection Point, and 2nd Derivative Test:P 195 (2,5,10,15,19,26,29,31,37)

Curve Sketching: P 196 (54,55,56)

Creating Equations with Initial Conditions:P 196 (61,62)

Section 3.5 – Limits at Infinity

Limits at Infinity

Limits at Infinity look at what happens to a function as its x values increase or decrease without bound.

Consider the function

Hopefully, you will notice that the limits as x approaches infinity and negative infinity mirror what you learned in PreCalculus about horizontal asymptotes.

2

2

3( )

1

xf x

x

x -∞← -100 -10 -1 0 1 10 100 →∞

f(x) 3← 2.9997 2.97 1.5 0 1.5 2.97 2.9997 →3

Rules for Finding Limits at Infinity1

lim 0

lim

0lim

0

lim

lim( ) lim lim

x

x

x

x

x x x

xc c

FindLimit

FindLimit

f g f g

lim

lim

lim

lim

lim

x

x

x

x

x

x

x c

x c

c x

x

c

Find the Limit at Infinity

2

4lim(3 )x x


3lim(x 2)x


3 1lim

1x

x

x

The trick with rational functions is to divide the top and bottom of the fraction by x to the largest power present.

Find the Limit at Infinity23 1

lim1x

x

x



2

3 1lim

1x

x

x


Limits at Infinity of Sin and Cossin

lim1x

x

x To help us determine this limit, let’s consider what we know about the graph of sin x

Limits at Infinity of Sin and Cos

limcosx

x

x x To help us determine this limit, let’s consider what we know about the graph of cos x

Limits at Infinity Homework

Limits at Infinity: P 205 (3,5,17,21-25,31-34)

Section 3.6 – Curve Sketching

Curve Sketching Without Calculators

We will use the following concepts to aid in sketching curves.•x and y intercepts•Vertical asymptotes•Increasing and decreasing•Relative extrema•Concavity•Points of Inflection•Horizontal Assymptotes•Limits at infinity

Sketching a polynomial

x intercepts:

y intercepts:

Vertical asymptotes:

Horizontal asymptotes:

4 3 2 3( ) 12 48 64 ( 4)f x x x x x x x

First Derivative:

Critical Numbers:

Second Derivative:

Points of Inflection:

4 3 2 3( ) 12 48 64 ( 4)f x x x x x x x

• Make an ascending list of all vertical asymptotes, critical numbers, and points of inflection.

• Make a chart compiling all the information

4 3 2 3( ) 12 48 64 ( 4)f x x x x x x x

x value or interval f(x) f’(x) f’’(x) Graph features

Sketching a Radical Function

x intercepts:

y intercepts:



5 43 3( ) 2 5f x x x

First Derivative:

Critical Numbers:

Second Derivative:


5 43 3( ) 2 5f x x x




5 43 3( ) 2 5f x x x

Sketching a Rational Function

x intercepts:

y intercepts:



2

2

2( 9)( )

4

xf x

x

First Derivative:

Critical Numbers:

Second Derivative:


2

2

2( 9)( )

4

xf x

x




2

2

2( 9)( )

4

xf x

x

Curve Sketching Homework

Sketching a Curve: P 215 (7,11,19,31,9*)

Section 3.7 – Optimization

Optimization

Optimization is the process of applying our understanding of relative maximums and minimums to word problems. In science, business, and product design the best and worst scenarios are found by setting the first derivative equal to zero.

Two numbers add up to 42. What is the most their product could be?We are going to use the idea that relative extrema are found where f’(x)=0 to solve this problem, but first we must write an equation.

First # = xSecond # = 42 – xTherefore theProduct = x (42-x) = 42x-x2

We can graph this function to get a feel for what it looks like.

Why did I choose Xmin=0 Xmax=42?

Let’s find the first derivative

Product = 42x-x2

Product’=42-2xWe find the relative extrema by setting Product’ = 0.0=42-2x2x=42x=21Remember the First # = x and the Second # = 42 – xFirst # = 21, Second # = 21, with product of 441.

We do want to be sure this is a maximum and not a minimum. Let’s use the 2nd derivative test.Product’’= -2So Product’’(21)=-2Consider our smiley face test. A negative 2nd derivative is frowning, concave down, and contains a maximum.

You are given 4 feet of wire, with which you are to make a square and circle. What size square and circle will result in the most combined area?Within the problem we find two equations.4 ft = (circumference of circle) + (perimeter of square)4 = 2πr + 4xArea = (area of circle) + (area of square)Area = πr2 + x2

Are there any restrictions on the value of x?

We need to substitute one equation into the other to eliminate a variable.4 = 2πr + 4xArea = πr2 + x2

4 2 4

4 4 2

4 4

22 2

r x

x r

xr

xr

2 22( )

2A ea x

xr

Let’s take a look at the 2nd derivative at x=1 to verify x=1 is a max.Area’’ = 2Area’’(1) = 2Since Area’’ is positive the face is smiling, the graph is concave up, and therefore x=1 is a minimum, instead of a maximum!!

Let’s take a look at the graph. Consider the restrictions…

On a closed interval where are the possible locations of the extrema?1.Critical Numbers2.Endpoints.

The best solution here is x = 0. The circle gets everything.

' 2 2

0 2 2

2 2

1

Area x

x

x

x

2

2

2 2

2 2

Area x x

Area x x

Optimization Homework

Optimizing: Worksheet

Documents

Chapter 3 – Spring 2014