sec ed 4646 final project

8/8/2019 Sec Ed 4646 Final Project

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12/10/2010

Sec Ed 4646 | Pettit, Amanda

UMSL FINAL PROJECT


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5.1: Composite Functions

Teacher: Pettit Subject: Mathematics, Math Analysis Level: 11/12th Grade

Topic: Composite Functions Lesson Duration: 1- 90 minute class

Advance Organizer: A composite function is a type of function that is made by inputting one function

into another.

Objectives:

1. After the lesson, the learner will be able to form a composite function. Math2. After the lesson, the learner will be able to find the domain of a composite function

Math A2B

Materials Needed:

Precalculus: Enhanced with Graphing Utilities, 5e. (2009). Sullivan & Sullivan. Pearson Prentice Hall

Paper, pencils, smart-board

Warm-up: 1) Find the domain of f(x) = ; 2) What is the definition of a function?

Phase 1: Introduction- Clarify aims and establish set

In the last four chapters, we have studied polynomials and rational functions, which are known

as algebraic functions. We can express these as sums, differences, products, quotients, powers and so

on. There are functions, however, that arent algebraic- these are called transcendental. They go

beyond algebraic functions.

Phase 2: Present Advanced Organizer

A composite function is a type of function that is made by inputting one function into another.

Phase 3: Present learning materials.

Present example in book as an introduction to the importance of composite functions.

Suppose there is a tanker leaking oil and your job as the engineer on the ship is to determine the area of

the circular oil patch around leak. It is determined that the oil is leaking from the tanker is such a way

that the radius of the circular patch of oil around the ship is increasing at a rate of 3 ft/min. Therefore,

the radius, r, of the oil patch at any time, t, in minutes, is given by r(t) = 3t. So after 20 minutes the

radius of the oil patch is r(20) = 3(20) = 60 ft!

The area, A, of a circle as a function of the radius r is given by A(r) = r2. The area around the area of the

circular patch of oil after 20 minutes is A(60) = (60)2 = 3600 square feet. Notice that 60 = r(20), so

A(60) = A(r(20)). The argument of the function A is itself a function! (a function within a function!).


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Generally, we can find the area of the oil patch as a function of time t by evaluating A(r(t)). This function

is a special type of function called a composite function.

(1) Define composite function.( Note: f g and f(g(x))

Are equivalent expressions)

Suppose we had two functions, f and g and x is in the domain of g. We can evaluate g at x and obtain

g(x). If g is in the domain of f, we can then evaluate g at g(x) and get f(g(x)) ( a function within a

function!) Essentially it is a function, composed of two functions. Use picture from book to illustrate.

Illustrations from text (p. 248-249)

Only the xs in the domain of for which g(x) is in the domain of f can be in the domain of f g. If g(x) isnt

in the domain of f, then we cant define f(g(x)). Use another picture to illustrate definition.

We input x into the function g to obtain g(x). We take the output g(x) and input it into f to get the

composite function f(g(x)).

Examples

I WORK (teacher): Let f(x) = 2x + 5 and g(x) = 3x2 Lets first find (f g)

f(g(x)) = 2(3x2) + 5 = 6x

2+ 5

Lets find (f g)(1) = f(g(1)). g(1) = 3(1) 2 = 3. f(3) = 2(3) + 5 = 11

NOW WORK (students): (f g)(5) Ans: (f g)(5) = 155


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Then have students work (g f)(5), (f f)(5) and (g g)(5) Ans: 675, 35, 16875

(2) Domain of Composite functionsThere are two steps to finding the domain of composite functions. Remember, we are dealing with two

functions that might not have the same domain.

Consider the two functions f(x) = 4 + 5 and g(x)= x 1. We know the domain of these two functions is

all real numbers or . Therefore the domain of f(g(x)) would be as well.

Now consider the two functions a(x) = and b(x) = . We all know these two functions have

different domains. How are we to determine the domain of their composition.

Step 1: Find the domain of the input function (the inside one). If there are any restrictions on it, keep

it!

Step 2: Construct the composite function. Find the domain of this new function. If there are restrictions

on the domain, add them to the restrictions from step 1. In case there is an overlap, use the more

restrictive domain (or their intersections). The new function may result in a domain unrelated to the

domains of the original functions!

Lets try an example.

I WORK: Let f(x) = x2

+ 2 and g(x) = .

Step 1: The domain of g(x) is . Keep this in mind!

Step 2: The composite function f(g(x)) = 5x is all reals, but we must keep the domain of the inside

function. So the domain for the composite function is also .

NOW WORK: Find the domain of g(f(x)).

Ans: Step 1: The domain of f(x) is all real numbers.

Step 2: The composite function g(f(x)) = . The domain of this new function is

. The input function f(x) has no restrictions so the domain of g(f(x)) is determined

only by the composite function. So the domain is .

Now have the students work a harder example where we have to consider two different domains.

I WORK: let f(x) = and g(x) = . Find (f g) and its domain.

Step 1: Find the domain of g(x).

Step 2: The composite function f(g(x)) = = = .


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The domain of this function is . Then putting the two together, we get a domain

.

Phase 4: Application

Now that we understand composite functions, we can use them to do useful things.

We can show that two composite functions are equal or (f g)(x) = (g f)(x) = x for every x in the domain

of (f g) and (g f).

I WORK: Suppose f(x) = 3x 4 and g(x) = (x + 4)

Lets find f(g(x)). f(g(x)) = 3( ) = x + 4 4 = x

Now, lets find g(f(x)) = [(3x -4) + 4] = (3x) = x

Therefore, we can conclude (f g)(x) = (g f)(x) = x !!!!!

Review material learned in class, if there is enough time, allow the students to start on their homework.

Homework Assignment: Worksheet 5.1 (attached)


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Worksheet 5.1 Composite Functions

Name:___________________________________ Date:___________

Math Analysis Section: 1 2 4

1. Evaluate the expression using the values given in the table.

(a) (f g)(1) (b) (f g)(-1) (c) (g f)(-1) (d) (f g)(0) (e) (g g)(-2)2. Evaluate the expression using the graphs of y = f(x) and y =g(x) in the figure.

(a) (g f)(1)(b) (g f)(5)(c) (f g)(0)(d) (f g)(2)

For questions 3 and 4, find (a) (f g)(4) (b) (g f)(2) (c) (f f)(1) (d) (g g)(0)

3. f(x) = 4x2 3, g(x) = 3 - x2 4. f(x) = , g(x) =

For 5 10, find the domain of the composite function (f g).

5. f(x) = , g(x) = 6. f(x) = x 2, g(x) = 7. f(x) = 2x + 3, g(x) = 3x

8. f(x) = - , g(x) = 9. f(x) = , g(x) = 10. f(x) = , g(x) =

For 11 12, show (f g)(x) = (g f)(x) = x for every x in the domain of (f g) and (g f).

11. f(x) = , g(x) = 12. f(x) = ax + b, g(x) = (x b), a


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13. If f and g are odd functions, show that the composite function (f g) is also odd.

14. If f is an odd function and g is an even function, show that the composite functions (f g) and (g f)

are both even.

15. The volume V of a right circular cone is V = r

3

. If the height is twice the radius, express the volumeV as a function of r.


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5.2: One- to- One & Inverse Functions


Topic: One- to- one & Inverse Functions Lesson Duration: 1- 90 minute class

Advance Organizer: If a function is one-to-one, then it has an inverse function .

Objectives:

1. After the lesson, the learner will be able to determine if a function is one-to-one2. After the lesson, the learner will be able to determine the inverse of a function defined by a map

or a set of ordered pairs

3. The learner will be able to graph the inverse function from the graph of the function after thelesson.

4. The learner will be able to find the inverse of a function defined by an equation after the lesson.Math A2B, Math G3B

Materials Needed:


Paper, pencils, smart-board, 1 Exploring Inverse Functions Student Worksheet for each student, and

one copy of Exploring Inverse Functions Teacher Notes worksheet.

Warm-up: 1) Explain in words what the vertical line test is and why its used. 2) Name 3 functions that

are either increasing or decreasing and list these such intervals (in interval notation!).


In previous chapters, we have learned to represent a function in four different ways: a map, a set of

ordered pairs, a graph and an equation. In this section we will learn about two special functions that are

vital to mathematics- the one-to-one function and the inverse function.

Phase 2: If a function is one-to-one, then it has an inverse function

Phase 3 Present learning materials

(1) determining if a function is one-to-oneFigure 1

Figure 2


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Figure 1 illustrates the correspondence between states and their populations (in millions). Figure 2

illustrates the correspondence between animals and life expectancy (in years). Suppose if I asked a

group of people to name a state that has a population of 0.8 million (Fig. 1). Everyone in the group

would say South Dakota. Now if we asked the same group to name an animal with a life expectancy of

11 years based on fig. 2, some would say a dog, while others would say a cat. The difference between

the two is that in figure 1, there are no two elements in the domain that correspond to the same

element in the range. But in figure 2, this isnt the case. Figure 2 is called a one-to-one function.

A function f is one-to-one if no y in the range is the image of more than one x in the domain. To

illustrate the relationship:

The figure (a) illustrates the distinction of a one-to-one function. (b) isnt one-to-one but is a function

and (c) isnt a function because x has two different images.

NOW WORK(students):

The below function illustrates the relationship between mens ages and HDL (bad cholestrol).

Is this a one-to-one function?

No, there are two different inputs, 55, and 61, that correspond

to the same output, 38.

Does represent a one-to-one function?

Yes because there are no two distinct inputs that correspond to the same output.


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The horizontal line y = 1

intersects the graph of

y = x^2 more than once.

So f isnt one-to-one.

The horizontal lines

only intersect the

graph of g exactly

once. g is one-to-

one.

Present Horizontal-line Test theorem

This test works because when we look at the figure to the left we see

where the horizontal line y = h intersects the graph at two places (x 1,

h) and (x2, h). h is in the image of both x1, and x2 and x1 x2.

f isnt one-to-one

In other words, if the graph of any horizontal line intersects the graph of the function f at more than onepoint, the function f isnt one-to-one.

NOW WORK: Have the students use the graph to determine whether or not the function is one-to-one.

(a) f(x) = (b) g(x) =

Can we make any generalizations? Present Increasing/Decreasing Theorem.

Lets look at g(x) = again. We know from previous sections that this function is increasing. An

increasing or decreasing function will always have different y-values for unequal x-values, we know a

function that is increasing or decreasing over its domain will be a one-to-one function.

(2) Inverse of a function defined by a map or a set of ordered pairs.Now we will discuss a correspondance called an inverse.

What is an inverse? We note an inverse by .


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The book states the following theorem

Consider the relationship seen earlier. We see that our x or state takes us to the output of population.

An inverse takes us back from the population to the state. In other

words, it undoes what f does. 6.2 takes us back to Indiana, 6.1

takes us back to Washington, 0.8 takes us back to South Dakota

and so on.

So, if the function f is a set of ordered pairs, say (x,y), then is the set of ordered pairs (y,x)

So the domain of f = range of .

The range of f = domain of .

3

Remember when we proved that (f g)(x) = (g f)(x) = x in the last section? This is true when we need to

verify inverse functions.

I WORK: Lets verify the inverse of g(x) = is (x) =

(g(x)) = ( = x

g( (x)) = g( = 3 = x

NOW WORK (students): Verify the inverse of f(x) = 2x + 3 is (x) = (x 3).

(3) Obtain a graph of an inverse function from the graph of the functionDefinition: Let f and g be two functions. If (f g)(x) = (g f)(x) = x, the g is the inverse of f and f is the

inverse of g.


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Ask students if they

can verbally explain

the relationship

between a function

and its inverse.

I WORK:Lets consider f(x) = + 2 and g(x) =

If we want to reflect something about the line y = x we get the points (a,b) and (b,a). Lets reflect a few

points about y = x

We see that the points are equidistant from each other about the line y = x. Now lets finish the above

example.

f is in blue and g is in green. We can easily see that the function

+ 2 is reflected about the line y = x.

Note: the reflected graph MUST pass the vertical line test to be

a function!

Definition: A function f is one-to-one IFF f has an inverse.

(4) Finding the inverse of a function defined by an equationSince we know that the graphs of a one-to-one function and its inverse are symmetric with respect to

y=x, we can get by interchanging the roles of x and y in f.

If f is defined by y = f(x) then is defined by x = f(y) . This defines implicitly, so if we can solve the

equation for y, we can get the explicit form of or y = .

There are three steps to finding the inverse of a one-to-one function.


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Replace f(x) with y.

Interchange the

variables x and y to get

x = f(y). This defines

the inverse function

implicitly.

I WORK: Lets find the inverse of f(x) = 2x + 3 and graph the function and its inverse on the same graph.

f(x) = 2x + 3 y = 2x + 3

x = 2y + 3

x = 2y + 3 2y + 3 = x 2y = x 3 y = (x 3)

Lets check the inverse by graphing the two functions and see if they are a reflection about the line y = x.

Note the symmetry about the line y = x.

NOW WORK (students): Find the inverse of f(x) = , x and check the result

Ans: (x) = , x .

Have students check the vertical asymptote of f, which is x = 1, and the horiztonal asymptote is y = 2.

The vertical asymptote of is x = 2, and the horizontal asymptote is y = 1. Curious

Phase 4: Application

We can also find the domain and range of inverse functions.

I WORK: Lets find the domain and range of f(x) =

The solve the implicit equation for y

in terms of x and get the explicit form

of the inverse y = (x)


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The domain of f is . To find the range of f, we can use the fact that the domain of equals

the range of f. So (x) = . The domain of is , so the range of f is . And

since the domain of f is , the range of is .

Now the students will do the Exploring Inverses Activity for the rest of the period. If they dont finish

it in class, it also part of their homework.

Attached: Exploring Inverse Functions Student Worksheet

Exploring Inverse Functions Teacher Notes

Assignment: 5.2 (p. 265) # 1-3, 7, 8, 9, 12, 15, 17-20 (yes or no), 24, 25, 28, 35, 39, 43, 44, 51, 53, 57, 63,

67, 69, 73, 76, 85, 95

Note: Some of these problems are yes or no or matching


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2009Texas Instruments Incorporated [email protected] 1.800.TI.CARES 1Getting Started with T I-Nspire Developmental Algebra

Name

Class

Problem StatementIn this activity, you will explore inverse relations and functions. Two relations are said to beinverses if they undo each other. For example, if a relation maps 5 to 2, then the inverserelation maps 2 back to 5. In general for a function, iff(a) b, then its inverse function isdenoted f1(b) a.

Inverse Point-by-Point

1. One way to find the inverse of a relation is to switch the mapping of thex- and y-coordinates of each ordered pair in the relation. That is, the inverse ofthe point (x, y) is (y,x) on the inverse relation. Find the inverse of this relation:{(2, 5), (4, 5), (9, 2), (0, 3)}.

2. Now that we have defined the inverse of a relation by the undoing nature it has forthe mapping of points, what does this definition mean when we think of the relationgraphically? On page 2.2 of the CollegeAlg_Inverses.tnsfile, you will find the functionf(x) x3 graphed and a point on the function labeled. Under the Construction menu,use the Measurement Transfertool to map thex-coordinate of the labeled point ontothe y-axis and the y-coordinate of the point onto thex-axis. To do this, click on thenumber in the ordered pair followed by the axis you desire.

Now plot the point on the inverse function by constructing a line perpendicular to thex-axis, passing through the point on thex-axis. Similarly, do the same for the point onthe y-axis. Now mark the intersection point of the two perpendicular lines. This is theinverse point of the point on the original function, f(x) x3.

Move the original point by dragging it. Describe how the inverse point moves.

Inverses as a Graphical Relation

3. On page 3.2, you will find the graph off(x) x2, along with point a on the graph. Theinverse of point a is also shown, created in the same way we did in the last question.On this graph, we have hidden the perpendicular lines so that we can observe themovement of the inverse point more easily.

a. Drag point a and describe how its inverse point moves.

b. Does the point move in a manner consistent with your observations from question 2?

Exploring Inverse Functions

Student Worksheet

Inverse_Functions_Student.indd 1Inverse_Functions_Student.indd 1 10/23/09 12:39:44 PM10/23/09 12:39:44 PM


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Developmental Algebra

2 2009Texas Instruments Incorporated Getting Started with TI-Nspire Developmental Algebra

c. To display the entire collection of inverse points, we can use the Locus tool foundunder the Construction menu. To do this, after selecting the Locus tool, click on theinverse point followed by the original point on the function. You will see the path ofthe inverse point appear. Describe the shape of the inverse relation. Does it matchyour earlier sketch? Sketch a path for your point below.

d. What relation does the locus of points appear to represent?

Inverses as Functions

4. In the previous problem, the graph of the inverse relation was not a function (evident bythe fact that it does not pass the vertical line test). However, the original relation was afunction (as it does pass the vertical line test).

a. Describe how you might test the graph of a relation to see if its inverse is a function.



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2009Texas Instruments Incorporated [email protected] 1.800.TI.CARES 3Getting Started with T I-Nspire Developmental Algebra

b. Decide if the inverse of the graph ofy x3 5 shown on page 4.4 of theCollegeAlg_Inverses.tnsfile is a function by checking for intersections with horizontallines. Display the locus of the inverse points to confirm your conclusion. Make asketch of your graph below.

Finding Inverses Algebraically

5. So far, this activity has focused on the method of switchingx- and y-values on a localor graphical scale (one point at a time) to find the inverse of a function. It is now time toswitch to a global scaleto algebraically calculate the inverse of a function. Here we willrequire that the inverse be a function as well.

To understand how to find the inverse function of a given function, we must first consideranother way that the inverse is understood. Recall from elementary school that foreven basic operations on numbers such as addition and multiplication, we have therelationship between inverse elements and the identity element. The identity elementis the element under the operation that leaves everything alone. For example, underaddition, 0 is the identity since 5 0 5, leaving 5 unchanged. Under multiplication,

1 acts as the identity since 7

1

7, leaving 7 unchanged. So what are the inversesin these cases? Consider that 5 5 0 and 7

1

__7 1. Here the idea is that if you

combine an element with its inverse, you get the identity element.

Exploring Inverse Functions

Student Worksheet



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Developmental Algebra

4 2009Texas Instruments Incorporated Getting Started with TI-Nspire Developmental Algebra

On the Calculatorpage 5.4, define a function, call it e(x), that you think will leaveall other functions unchanged when composed with them. Then test your identityfunction by composing it with other functions you define. Do your other functionsremain unchanged? Continue until you find an identity function. Is your identity functionsurprising to you? Explain.

6. The idea of an inverse is to get the identity function when it is composed with the originalfunction. Based on your observation from the last question, this would mean thatf(f1(x)) x, where f1(x) represents the inverse function off(x). On the Calculatorpage 6.2, define f(x) x3 5. Now to find the inverse function, we would like f(y) x,where yhere is representing the inverse function (note the switching of the x- and

y-coordinates as we did in the graphical swap in question 2).

a. Use the solve( command under the Algebra menu to solve the equation f(y) xfory.The syntax for the solve( command is solve(f(y)=x,y). Give your inverse function andgo back to page 4.4 and graph it. Does it match the locus you created? Sketch thegraph of your inverse function below.

b. Now go back to page 6.2 and define your new inverse function as g(x). Composef(g(x)). Do you get the expected identity function? Explain why your compositionresult should yield justx.



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Ex plor ing Inverse Func t ionsTeacher Notes

2009 Texas Instruments Page 1 Exploring Inverse Functions

Activity OverviewStudents will investigate the fundamental concept of an inverse, generate the inverse graphs ofrelations applying this concept, and algebraically determine the inverse.

Materials Technology:TI-Nspire handheld, TI-Nspire CAS handheld, or TI-Nspire CAS computer

software

Documents:Inverse_Functions.tns, Inverse_Functions_Student.doc

Student Solutions

Inverses Point-by-Point

1. {(5, 2), (5, 4), (2, 9), (3, 0)}2. Point moves like the original function, only flipped about y= x.

Inverses as a Graphical Relation

3. a. Moves like the original parabola turned on its side

b. Yes

c.

d. Two halves of y x= and y x= or simply y2 = x

Inverses as Functions

4. a. If there exists a horizontal line that intersects the graph at more than one point, then the

inverse relation is not a function.


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Ex plor ing Inverse Func t ionsTeacher Notes

2009 Texas Instruments Page 2 Exploring Inverse Functions

b. It is a function.

Finding Inverses Algebraically

5.

6. a. Yes, it matches the locus.

b. The composition gives a result of x. This is important because the composition of a

function with its inverse should give the identity function (maps xdirectly back to x).


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5.3: Exponential Functions


Topic: Exponential Functions Lesson Duration: 2- 90 minute classes

Advance Organizer: Exponential functions are vital to mathematics and they have many uniqueproperties.

Objectives:

1. After the lesson, the learner will be able to evaluate and graph exponential functions.2. After the lesson, the learner will be able define e.3. The learner will be able to solve exponential equations after the lesson.

Math A2B, A1E, A1D, A2C, G3B,

Materials Needed:

Precalculus: Enhanced with Graphing Utilities, 5e. (2009). Sullivan & Sullivan. Pearson Prentice Hall.

Paper, pencils, smart-board

Warm-up: 1) Graph the equation 2x + 3 using transformations.

2) Find the horizontal asymptotes of R(x) =


In earlier part of the course, we have discussed raising a real number a to a rational power and thus we

gave meaning to the form where a is a positive real number and r is some rational number. But what

happens when we consider the case where x is an irrational number?

Phase 2: Present advanced organizer: Exponential functions are vital to mathematics and they have

many unique properties.

Phase 3: Present learning materials

(1) Evaluation exponential functionsAlthough we cant truly define when r is an irrational number, we can present a basic definition. We can

choose a rational number r that is made by removing all but a finite number of digits from the irrational

number x. We can expect that . Suppose we take = 3.14159. Then we can approximate

Lets learn some properties for real exponents that we know from rational exponents.


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Ask the students:

What do we notice

about the

relationship

between x and

f(x)?

We double the value

of f at 0 to get the

value at 1. Also we

double the value at 1

to get the value at 2

We have been introduced to such laws in previous math courses, so the laws for real exponents are no

different.

NOW WORK (students): Have the students evaluate the following expressions with their graphing

calculators.

(a) (b) (c) (d) (e)Ans: (a) 2.639015822 (b) 2.657371628 (c) 2.66474965 (d) 2.665119089 (e) 2.665144143

NOW WORK: Tell the students we need to find a function f that

(1) The value of f doubles with every 1-unit increase in the x direction(2) The value of f at x = 0 is 5, so f(0) = 5

Help the students construct the table

x f(x)0 5

1 10

2 20

3 40

4 80

(The key fact is that the value of f doubles for every 1-unit increase in x)

f(0) = 5

f(1) = 2f(0) = 2 * 5 = 5 *

f(2) = 2f(1) = 2(5 * 2) = 5 *

f(3) = 2f(2) = 2(5 * ) = 5 *

f(4) = 2f(3) = 2(5 * ) = 5 *

The pattern is: f(x) = 2f(x 1) = 2(5 * ) = 5 *

Lead into

Why a 1? We exclude a = 1 because its the constant function C * = C and we also exclude negative

bases. A few examples of exponential functions are f(x) = and ( )X


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Its increasing so

therefore its a

one-to-one

function.

Theorem

Lets prove this!

= = = =

(3) Graphing Exponential FunctionsLets first learn how to graph the exponential function f(x) = , then perhaps we can do some

transformations to get the graph of f(x) = C *

NOW WORK: Have the students graph f(x) = on their graphing calculators. Ask them what the

domain is (). Ask them about x intercepts (there are none). The whole graph lies above the x-axis

for all x. The graph has an asymptote at y = 0. Ask the students as x approaches - , the value of f getscloser to 0.

( The above is what should appear on their graphing utilities).

As x approaches , f(x) = grows very quickly, causing the graph to rise

sharply. It is an increasing function, so what can we say about the function?

We see this kind of behavior in many situations, like population growth,

bacterial growth etc. We call this exponential growth.

Lets learn about properties of Exponential Functions


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The domain is the set of real numbers and the range is the set of positive real numbers.

I WORK: Lets graph f(x) = (

We know the domain to be all real numbers, and we therefore can get some points of the graph.

Since > 0 for all x, the range is (0, . It lies above the x-axis and

therefore has no x-intercepts. The y-intercept is y = 1 and as x

approaches - , f grows rapidly and as x approaches the values of f

approaches 0.

We could gotten the graph of f(x) = ( from the

graph of f(x) = . The graph of f(x) = ( = . Its a reflection aboutthe y-axis.

Lets look at the graph of two different exponential functions

The graphs of f(x) = ( and f(x) = ( . The bases

are between 0 and 1. Notice, the smaller the base, the

steeper the graph becomes when x < 0, and when x > 0,

the smaller the base, the closer the graph of thefunction is to the x-axis.

Lead into below table of properties..

Is the graph of f

increasing or

decreasing? (its

decreasing, so one-

to-one).


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Then replace x by

x and reflect

about the y-axis

Graph of y = . Then subtract 3 and

shift down 3 units.

Remember when I mentioned we can graph exponential functions using transformations?

NOW WORK: Have students do the example with you.

Lets graph f(x) = 3 and find the domain, range, and horizontal asymptotes.

Lets start with the graph of y = .

What is the domain? (- , ). Range? (-3, ) Horizontal Asymptote? (y = -3)

(2) The number e.What is e? It

s a number like

. E = 2.718281828459045235306

, where

=3.14159265358979323846 Like pi, e never ends.

We can define it by

We can create a table of values for when n takes on increasingly large values. The last digit is correct to

9 decimal places.


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The exponential function f(x) = , whose base is e is very important in mathematics because it occurs

very frequently it is considered THE exponential function.

The graph of the function is as illustrated. Since 2 < e < 3, its an increasing function and lies between

the graphs of y = and y = .

We can graph y = using transformations!

I WORK: Lets graph f(x) = - and find the domain, range, and asymptote.

First (as always), lets start with the parent function y =

The graph of y = Multiply by -1;

Reflect about

x-axis.

Replace x by x 3;

Shift right 3 units.


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The domain is (- , ). Range is (- , 0). The horizontal asymptote is y = 0.

(3) Solving exponential equationsWe can solve exponential equations by applying the Laws of Exponents and the following property

The above property is a consequence of the fact that exponential functions are one-to-one. A further

note is that each side of the equality MUST have the same base to be able to use the above property.

I WORK: Lets solve = 81

First lets do it algebraically

= 81 (81 = ). Now that we have the same base on each side, we can set the

exponents equal to each other and solve. x + 1 = 4 x = 3. The solution set is { 3 }.

Now, lets do it graphically. Graph each equation on your graphing calculator. Let Y1 = and Y2= 81.

The graph intersects at (3, 81), so the solution set is { 3 }.

NOW WORK (students): Solve = )2*

Use the law of exponents to get the same base of e on both sides.

)2

* = * = As a result, = 2x -3 x = -3 or 1. So the solution

set is { -3, 1}. Have the students then verify their answer on their graphing calculators.


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Homework: (p. 279) 1, 5, 6, 11, 14, 21, 27, 29, 32, 34, 41, 47, 49, 61, 72, 78, 83, 95, 108 (a-c only), 111,

120, 121.


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Name: ______________________________________ Date:_______________

Math Analysis: Quiz 5.1- 5.3 Period: 1 2 4

Directions: Complete the following questions. To receive full credit, show all work and make your

answers clear!

1. Evaluate the expression using the graph of y = f(x) and y= g(x) shown below.(a) (g f)(-1) = __________(b) (g f)(0) = __________(c) (f g)(-1) = __________(d) (f g)(4) = __________

2. Find the composite function (f g) and its domain.f(x) = ; g(x) =

3. Show that (f g)(x) = (g f)(x) = x for every x in the domain of (f g) and (g f).f(x) = 2x 6 g(x) = (x + 6)

4. Is every odd function one-to-one? Explain.

5. A function f has an inverse function. If the graph of f lies in quadrant II, in which quadrant doesthe graph of lie? Why?

6.

If the graph of the exponential function f(x) = , a > 0, a

1, is decreasing, then a must be lessthan _______.

7. Graph the function and determine the domain and range.

f(x) =


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8. Solve

9. =

10. If = 3, what does equal?


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Quiz 5.1- 5.3 Self-AssessmentName ____________________________________ Math Analysis Period: 1 2 4

As you review each problem on your test, mark whether each problem is right or wrong in the table below withan X. Then look at all of the questions you answered incorrectly, decide which of these were due to simple

mistakes, and mark the Simple Mistake column with an X. For all remaining incorrect answers, mark the

Dont Get It column for that question.

Question#

Learning Target Right? Wrong? SimpleMistake?

Dont

Get

It

1 Evaluating composite functions.

2 Forming a composite function and finding its

domain.

3 Proving a two functions are inverses.

4 The ability to determine whether or not a functionis a one-to-one function.

5 Understand properties of inverses.

6 Understanding the properties of exponentialfunctions.

7 Ability to graph an exponential function

8 Ability to solve an exponential equation. Using

the laws of exponents and property (3).

9 Ability to solve an exponential equation. Using

the laws of exponents and property (3).

10 Ability to solve an exponential equation using the

laws of exponents and property (3).


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To identify your areas of strength, write down the learning target corresponding to the problems you feltconfident about and got right. Then write a short description of the problem.

Learning

Target #

Learning Target or Problem Description

To determine what you need to study most, write down the learning target corresponding to the marks in the

Dont Get It column (problems you answered incorrectly NOT because of a simple mistake). Then write a

short description of the problem.

LearningTarget #


To determine where you make your simple mistakes, write down the learning target or problem description for

those questions you marked in the Simple Mistake column and also include a description of your mistake.

Learning

Target #



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5.4: Logarithmic Functions


Topic: Logarithmic Functions Lesson Duration: 2- 90 minute class

Advance Organizer: Logarithmic functions and exponential functions are inverses.

Objectives:

1. After the lesson, the learner will be able to change exponential statements to logarithmicstatements and vice-versa.

2. After the lesson, the learner will be able to evaluate logarithmic statements.3. The learner will be able to find the domain of the logarithmic function after the lesson.4. The learner will be able to graph a logarithmic function after the lesson.5. After the lesson, the student will be able to solve logarithmic equations.

Math G3B, G4B, A2C, A4A, A1D, A1E

Materials Needed:


Paper, pencils, smart-board, 1 copy of Logarithmic Function Activity Worksheet (Explore Math) for each

student.

Warm-up: 1) Solve 2 < x + 10 and graph the solution set. 2) Solve > x and graph the solution set.


Two sections ago we learned a one-to-one function has an inverse and can be defined implicitly

by x = f(y). We also learned last section the exponential function is one-to-one and has an inverse

(implicit) x = . This function is very important in mathematics and is what is called a logarithmic

function.

Phase 2: Present Advanced Organizer: Logarithmic functions and exponential functions are inverses.

Phase 3: Present learning materials.

Lets start with a theorem.


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A logarithm is a name for a certain exponent!

Relate prior knowledge.

I WORK: If y = , then x = For example, 4 = is equivalent to 81 = .

If y = , then x = . For example, -1 = is equivalent to = .

(1) Change exponential statements to logarithmic statements and vice-versaLets change an exponential statement to a logarithmic one.

If = m, then 3 = If = 9, then b =

Now, lets go the other way

If = 5, then = 4 If = -3, then = b

(2) Evaluate Logarithmic expressionsIn order to find the exact value of a logarithm, we need to use exponential notation and use the

property that if = , then u = v. (Recall last section!)

I WORK: Lets find the exact value of

y = = 16 = y = 4

NOW WORK (students): Ans: y = -3

Recap: we need to make sure our bases are THE SAME! This will never change and dont even think

about trying to solve logarithms without using this fact!

(3) Domain of a Logarithmic FunctionSo far we have defined y = as the inverse of y = . Recalling section 5.2, we learned that

Domain of = Range of f Range of = Domain of f

Naturally, we can see

Now, it is a proper time to define some properties of logarithmic functions


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The domain consists of POSITIVE real numbers and therefore the argument (say x) of a logarithmic

function must always be greater than 0.

I WORK: Lets find the domain of f(x) =

The domain { x | x > -3} or (-1, 1).

NOW WORK: Find the domain of a) b) p(x) =

Ans: a) (-1, 1) (b) (-, 0)U(0, )

(4) Graphing Log FunctionsWe know from above that the exponential and logarithmic functions are inverses, so therefore we

know that the graph of the log function is a reflection about the line y = x of the graph of the

exponential function.

Lets see an example.

To graph y = , we need to first graph y = and then reflect it

about the line y = x.

NOW WORK: Graph f(x) =

Ans: graph and reflect it about y = x

For 0 < a < 1 For a > 1


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Now introduce properties of Log functions

If the base of our log function is e, then we have what is known as the natural logarithm function. It is

very frequent in mathematics and is given the symbol ln (stands for logarithmus naturalis).

y = ln x and y = are inverse functions and therefore we can get the

graph of y = ln x by reflecting the graph of y = about y = x.

NOW WORK: Have the students graph f(x) = ln x on their graphing

calculators. Have them access the table function on their

calculators. Notice for x 0, we get an error message. Why?

NOW WORK: Have the students pair up with a class- mate to answer the following questions.

Consider f(x) = - In(x 2)

a) Find domainb) Graph fc) Determine the range, and the vertical asymptote.d) Find the inverse of fe) Use inverse to confirm the range of f (found in c). From the domain of f, find the range of the

inverse.

f) Graph the inverse.Ans. a) (2, )

We can say y = is

equivalent to y = ln x


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b) c) Range is the set of all reals, vertical asymptote

is at x = 2 (Make sure they tell you the original

asymptote was shifted right by 2)

d) = + 2

e) (2, )

f)

If the base of the log function is 10, then we have what

is known as the common log function. If it isnt

indicated, its assumed to be 10

y = IFF x =

(5) Solving Logarithmic functionsWe must be careful when solving logarithmic equations because in the expression , a and x are

positive and a 1. We can sometimes solve a logarithmic function by changing it from a logarithmic

function to an exponential one.


= 2 4x 7 = 4x 7 = 9 x = 4. The solution set { 4 }

We can check ourselves = = = 2

NOW WORK: = 2 Ans: The solution set is { 8 }

We can also use logarithms to solve exponential equations


= 5 In 5 = 2x x = 0.805. The solution set is { }

Try to keep the solution as exact as you can.


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To summarize, draw attention to blue box on p. 293

Do Activity: hand-out ExploreMath Logarithmic Functions Activity to students. Try to get them to finish

them in class, if not, they can finish at home.

Homework: 1, 3 14, 16, 24, 31, 47, 59, 63-70, 79, 85, 103, 134, 135


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ExploreMath.com Lesson Plan>>Logarithmic Functions (Worksheet Version)>>Page 1 of 3

Worksheet

Logarithmic Functions

Introduction

Logarithmic functions are the inverses of exponential functions. Logarithms arefound in applications such as compound interest, earthquake magnitudes, andthe pH of solutions. In this activity we will look at logarithmic functions in the formy = logax.

Lets explore logarithmic functions. Use your web browser to go to theLogarithms activity:

http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=7

Below are several questions designed to get you thinking about the activity.Answer them on a separate sheet of paper.

Defining y = logax

Grab the a slide bar and move it such that a=4. This can also be accomplishedby typing a 4 to the right of the slide bar.

Now select the Calculate Data Values clipboard at the bottom of the screen.Then set the minimum to 0, set the maximum to 64 and set the step to 1.
http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=7http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=7http://www.exploremath.com/activities/Activity_page.cfm?ActivityID=7


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ExploreMath.com Lesson Plan>>Logarithmic Functions (Worksheet Version)>>Page 2 of 3

Pick out the x, y pairs that have integer values. These should be:

x 1 4 16 64y 0 1 2 3

Remember that the base, a, of this logarithm is 4.

Question 1a. How does a, y, and x related to each other?

Question 1b. What conjecture can you make about the definition of y = logax?

Question 1c. What is the value of y in the equation y = log381?

Restrictions of logarithmic functions

Grab the a slide bar and slide it to the left and right.

Question 2a. What happens to the graph when x approaches the y-axis?

Question 2b. Does the graph ever cross the y-axis?

Question 2c. Why do the graphs never enter the second or third quadrants?Remember that ay=x.

Question 2d. Why is a always positive? Think about inverses of exponentialfunctions.

Question 2e. Why cant a =1 in a logarithmic function? Set a = 1 and observe thegraph.


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Question 2f. What point does the graphs of logarithmic functions pass throughregardless of the base? Experiment with the a slide bar to help find the answer.

Algebraic inverses

Type e5 into your calculator. Now take the ln of the answer.

Question 3a. What do you notice about the final answer compared to theoriginal expression?

Question 3b. What would be the solution of log8(85)? Why?

Question 3c. What would be the solution of e(ln9)? Why?

Applications

Question 4a. If $200 were invested in an account where the interest iscompounded continuously at a rate of 5%, the amount, A, in the account after tyears could be evaluated using the equation: A = 200e.05t. How long would ittake the account to accumulate to $500?

The pH of a chemical is given by the formula: PH = -log10[H+], where [H+] is the

concentration of hydrogen ions in moles per liter.

Question 4b. What is the hydrogen ion concentration of an acid with a pH=4.2?

Conclusion

Logarithmic functions are the inverses of exponential functions. A logarithm isdefined as: y = logax if and only if x = a

y. The domain of a logarithmic function islimited to positive values of x Logarithms are used in a variety of applications