Lesson 1: Functions as Models

A function happen to be a simple mathemical model or a piece of larger model.

Functions describe situations where one quantity determines another. For ex-

ample, the return on P10,000 invested at an annualized percentage rate of 4.25

is a function of the length of time the money is invested. Because we continu-

ally make theories about dependencies between quantities in nature and society,

functions are important tools in the construction of mathematical models.

In school mathematics, functions usually have numerical inputs and outputs

and are often dened by an algebraic expression. For example, the time in

hours it takes for a car to drive 100 miles is a function of the cars speed in miles

per hour, v; the rule T(v) = 100/v expresses this relationship algebraically and

denes a function whose name is T.

The set of inputs to a function is called its domain. We often infer the do-

main to be all inputs for which the expression dening a function has a value,

or for which the function makes sense in a given context.

A function can be described in various ways, such as by a graph (e.g., the

trace of a seismograph); by a verbal rule, as in, Ill give you a state, you give me

the capital city; by an algebraic expression like f(x) = a + bx; or by a recursive

rule. The graph of a function is often a useful way of visualizing the relation-

ship of the function models, and manipulating a mathematical expression for a

function can throw light on the functions properties.

Functions presented as expressions can model many important phenomena. Two

important families of functions characterized by laws of growth are linear func-

tions, which grow at a constant rate, and exponential functions, which grow at

a constant percent rate. Linear functions with a constant term of zero describe

proportional relationships.

A graphing utility or a computer algebra system can be used to experiment

with properties of these functions and their graphs and to build computational

models of functions, including recursively dened functions.

Example 1: The graph of a function h is shown below:

(a) Find the values h(1) and h(5).

(b) What is the domain and range of h?

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Solution:

(a) We see from the graph that the point (1,3) lies on the graph of h, so the

value of h at 1 is h(1) = 3. While x = 5, the graph lies about 0.7 unit below

the x- axis. Therefore, we estimate that h(5) = -0.7.

(b) Notice that h(x) is dened when 0 x 7, so the domain of f is the

closed interval 0,7. See that h takes on all values from the interval -2 to 4, so

the range of h is -2 y 4 = -2,4

Example 2: Sketch the graph and nd the domain and range of the func-

tions:

(a) f(x) = 2x- 1

(b) g(x) = x2

Solution:

(a)The equation of graph is y = 2x-1, and recognize this as being the equa-

tion of a line with slope 2 and y-intercept -1.

Recall: The slope- intercept form of the equation of a line y = mx + b.

This enables us to sketch the graph below. The expression is dened for all

real numbers, so the domain of is the set of all real numbers, which we denote

by R. The graph shows that the range is also R.

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(b) Since g(2) = 2

2

= 4 and g(-1) = ( 1) 2

= 1, we could plot the points (2,4)

and (-1,1), together with a few other points on the graph, and join them to pro-

duce the graph below. The equation of the graph is y = x2

, which represents a

parabola. The domain of g is R. The range of g consists of all values of g(x),

and that is the all numbers of the form x2

. But x2

0 for all numbers x and

any positive number of y is a square. Therefore, the range of g is y| y 0 =

0, 1). This can also be seen from the gure below. Example 3: The cost of a pound of orange juice for three consecutuve week

is given by the table below:

The Price of Orange Juice Week Week 1 Week 2 Week 3

Cost 200 215 230

What will be the cost of a pound of orange juice be in Week 4?

Solution: The actual cost of a pound of orange juice in Week 4 will be de-

termined by a number of factors, such as orange juice production, distribution,

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sales, etc. These factors are the natural law governing orange juice cost. The

recent cost of orange juice can be model as:

x= the number of weeks since Week 1

P(x) = the cost of a pound of orange juice at time x, in pesos

The table above have shown us the details: P(0) = 200, P(1) = 215, and P(2) =

230. Then summarizing this information with the function will show us: P(x)

= 200 + (15)x.

Using the model, we can deduce that P(3) = 200 +(15)(3) = 245 pesos. We can

now predict that the cost of a pound of orange juice in Week 4 will be 245 pesos.

Note that this may or may not be accurate. The model between the relation-

ship of orange juice and it’s cost is based entirely on an observation of previous

patterns.

Example 4: The populations P

B and

P

M of two colonies of penguins, along

the Broughton and Montague Island of the New South Wales, respectively, have

been successfully modeled by the following two functions:

PB (t) = 2+(0.02)

t3

,

P M (t) = (1

:1) t

where the populations are measured in thousands of pairs of penguins and t is

measured in years from the present. When are the two colonies of equal size?

Solution: Since both of the functions are given by algebraic representations

formulas a natural instinct might be to try to solve the problem using an alge-

braic method. In algebraic terms, the problem would be to solve the following

equation for t :

2 + (0.002)t3

= (1 :1) t

.

However, this equation does not have an algebraic solution! (And it isnt because

we just dont know how to solve it.) Algebraic methods of solution failed.

Does this mean that the problem has no solution?

Before abandoning the algebraic approach entirely, it will serve as a clue, if

not a complete solution. Notice that right now (t = 0)

PB (0) = 2+(0.002)(0) 3

= 2,

P M (0) = (1

:1) 0

= 1.

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The Broughton colony currently consists of 2000 penguins to the Montague

colonys 1000. The only way the Broughton’s and Montague’s populations will

ever be equal at some point in the future is if the Montague colony outgrows

the Broughton colony. See that we must look at the long-term behavior of the

functions. It is exactly this kind of big picture for which another method – the

graphical method – is best suited.

Plot the two population functions and convert the algebraic representations

to graphical ones. Over the next 36 years, the plots are shown below: The top blue curve represents the Broughton colony. It appears that the

Broughton colony not only has a head start (observed algebraically above, but

not so apparent here), but it is also rapidly outgrowing the Montague colony.

This would seem to settle things: The Montague colony will never equal the

size of the Broughton colony. Unfortunately, this is the wrong.

While graphical methods are able to give the big picture of a functions behavior,

there is always the question of how big is big enough. If we had looked at the

populations for a period of time twice as long, we would have seen the following: In this view it is apparent that the Montague population does catch up with

the Broughton population, somewhere between 60 and 70 years from now. We

could now use the zooming feature on our calculator or computer to nd the in-

tersection point and get a better estimate of the time when the two populations

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will be the same.

Have we found all of the solutions? You should, at least, be skeptical by now.

Perhaps a still larger viewing window would reveal that the Broughton popu-

lation eventually retakes the Montague population. Extending the time scale

farther and farther into the future, however, shows no such trend. (What does

it show?) After a bit of experimentation in this direction, we may be ready to

conclude that the populations are equal only once. Again, we would be wrong.

If we extend our models only a short time into the past, we see that the popu-

lations were also equal a little less than ten years ago. The plots look like this: Example 5: New South Wales Penguins

See previous example

Solution: Notice that the algebraic problem of solving 2 + (0.002)t3

= (1 :1) t

(abandoned previously) is the same as solving 2 + (0.002)t3

– (1 :1) t

= 0

That is the same as nding the root of the function f(t) = 2 + (0.002) t3

– (1 :1) t

.

Using the graphical observation that the populations appear to be equal at some

point between t = 60 and t = 70, we might use the formula for f to calculate

approximate values of f(60) and f(70):

First Iteration t 60 70

f(t) 121.52 -101.75

Since the value of the function changes sign between t = 60 and t = 70, we have

conrmed that a root lies at some point in-between. Suppose we look half way

in-between:

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Second Iretation

t 60 65 70

f(t) 121.52 60.88 -101.75

Then we see that the value of f changes sign between t = 65 and t = 70, and

therefore the root must lie at some point in this interval. Computing the value

half way across this new interval leads to another table entry:

Third Iteration t 60 65 67.5 70

f(t) 121.52 60.88 -5.22 -101.75

Then notice that the root is in the interval between t = 67.5 and t = 70.

The method is clear enough to keep subdividing the one interval on which f

changes sign and compute a new value for the table at the midpoint. Even-

tually, after many iterations of this simple step, we will be able to compute

the root to any degree of accuracy. This may seem tedious, but its algorithmic

nature makes it an ideal method for use on calculators or computers, which can

perform tedious calculations very quickly.

After ve more iterations, the root is bracketed between t = 67.27 and t =

67.34. We therefore, conclude that, to one decimal place, the root is at t =

67.3. This much accuracy could have been achieved fairly quickly by graphing

and zooming. Unlike graphical methods, however, there is no limit to the degree

of accuracy that we may obtain by using the numerical method over and over

again.

Conclusion: Good mathematical models use the strengths of one representa-

tion to make up for the weaknesses of another. Good mathematical modelers,

likewise, use the strengths of one method to make up for the anothers weak-

nesses. Whether we are modeling a simple cause and eect relationship or a

complex physical system, we must look at problems from a variety of points of

view, and make use of all of the tools that are available to us.

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Lesson 2: Evaluating Functions

To evaluate a function

1.Substitute the given value in the function of x.

2.Replace all the variable xwith the value of the function.

3.Then compute and simplify the given function.

Example 1: Given the function: f(x ) = 2 x+ 1, nd f(6).

Substitute 6 in place holder x,

f(6) = 2 x+ 1

Replace all the variable of xwith 6,

f(6) = 2(6) + 1

Then compute function. f(6) = 12 + 1

f (6) = 13

Therefore, f(6) = 13. It can also write in ordered pair (6,13).

Example 2: Given the function f(x ) = x2

+ 2 x+ 4 when x= 4. Substitute

-4 in the place holder x,

f( 4) = x2

+ 2 x+ 4

Replace the all the variables with 6, f( 4) = ( 4) 2

+ 2( 4) + 4

f ( 4) = (16) + ( 8) + 4

f ( 4) = 12

Therefore, f( 4) = 12 or simply as ( 4;12) :

Example 3: Given g(x ) = x2

+ 2 x- 1. Find g(2y).

Answer in terms of y.

g(2 y) = x2

+ 2 x 1

g (2 y) = (2 y)2

+ 2(2 y) 1

g (2 y) = 4 y2

+ 4 y 1

Therefore, 4( y)2

+ 4 y 1:

8

Example 4: Given

f(x ) = 2 x2

+ 4 x- 12, nd f(2 x+ 4).

Solution:

f(2 x+ 4) = 2 x2

+ 4 x 12

= 2(2 x+ 4) 2

+ 4(2 x+ 4) 12

= 2(2 x+ 4)(2 x+ 4) + 4(2 x+ 4) 12

= 2(4 x2

+ 16 x+ 16) + 4(2 x+ 4) 12

= (8 x2

+ 32 x+ 32) + (8 x+ 16) 12

Combine like terms f(2 x+ 4) = 8 x2

+ (32 x+ 8 x) + (32 + 16 12)

= 8 x2

+ 40 x+ 36

= 2(2 x2

+ 10 x+ 9)

Therefore, f(2 x+ 4) = 2(2 x2

+ 10 x+ 9).

Example 5: Given f(x ) = x2

-x – 4. If f(m ) = 8, compute the value of m

Solution: Make the function f(x ) equivalent to f(m )

x 2

x 4 = 8

x 2

x 12 = 0

( x 4)( x+ 3) = 0

x 4 + 0; x+ 3 = 0

x = 4; x= 3

Therefore, the value of a can be either 4 or -3.

9

Exercises:

Evaluate the functions

given:

1. p(x ) = 2x + 1, nd p(-2)

2. p(x ) = 4 x, nd p(-4)

3. g(n ) = 3 n2

+ 6, nd g(8)

4. g(x ) = x3

+ 4 x, nd g(5)

5. f(n ) = n3

+ 3 n2

, nd f(-5)

6. w(a ) = a2

+ 5 a, nd w(7)

7. p(a ) = a3

– 4 a, nd p(-6)

8. f(n ) = 4 3

n

+ 8 5

, nd

f(-1)

9. f(x) = -1 + 1 4

x;

nd f(3 4

)

10. h(n) = n3

+ 6 n, nd h(4)

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Answers in Exercises:

1. 5

2. -16

3. 198

4. 145

5. -50

6. 84

7. -192

8. 4 15

9. – 13 16

10. 88

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Lesson 3: Operations on Functions

Let h(x) and g(x) be functions, and the operations on these two functions is

shown below:

Adding two functions as:

(h+g)(x) = h(x)+g(x)

Subtracting two functions as:

(h-g)(x) = h(x) – g(x)

Multiplying two functions as:

(h g)(x) = h(x) g(c)

Dividing two functions as:

( h g

)(x) = h

(x ) g

(x ) ; whereg

(x ) 6

= 0

Example 1:

Let f(x) = 4x + 5 and g(x) = 3x. Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g

)(x).

(f+g)(x) = (4x+5) + (3x) = 7x+5

(f-g)(x) = (4x+5) – (3x) = x+5

(f g)(x) = (4x+5) (3x) = 12 x2

+5x

(f g

)(x) = 4

x +5 3

x

Example 2:

Let f(x)= 3x+2 and g(x)= 5x-1. Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g

)(x).

(f+g)(x) = (3x+2) + (5x-1) = 8x+1

(f-g)(x) = (3x+2) – (5x-1) = -2x+3

(f g) = (3x+2) (5x-1) = 15 x2

+7x -2

(f g

)(x) = 3

x +2 5

x 1

Example 3:

Let v(x) = x3

and w(x) = 3 x2

+5x. Find (v+w)(x), (v-w)(x), (v w)(x), and

( v w

)(x).

(v+w)(x) = ( x3

) + (3 x2

+5x) = x3

+ 3 x2

+5x

(v-w)(x) = ( x3

) (3×2

+5x) = x3

3x 2

-5x

(v w) = ( x3

) (3×2

+5x) = 3 x5

+ 5 x4

(v w

)(x) = ( x

3 3

x 2

+5 x) = x

x 2 x

(3 x+5) = x

2 3

x +5

Example 4:

Let f(x) = 4 x3

+ 2 x2

+4x + 1 and g(x) = 3 x5

+ 4 x2

+8x-12. Find (f+g)(x),

(f-g)(x), (f g)(x), and ( f g

)(x).

12

(f+g)(x) = (4 x3

+ 2 x2

+4x+1) + (3 x5

+ 4 x2

+8x-12) = 3 x5

+ 4 x3

+ 6 x2

+12x

-11

(f-g)(x) = (4 x3

+ 2 x2

+4x+1) – (3 x5

+ 4 x2

+8x-12) = 3x 5

+ 4 x3

2x 2

-4x+13

(f g)(x) = (4 x3

+ 2 x2

+4x+1) (3 x5

+ 4 x2

+8x-12)

= 12 x8

+ 6 x7

+ 12 x6

+ 19 x5

+ 40 x4

16×3

+ 12 x2

40x 12

(f g

)(x) = (4

x3

+2 x2

+4 x+1) (3

x5

+4 x2

+8 x 12)

Example 5:

Let h(x) = 1 and g(x) = x4

x3

+ x2

-1. Find (h+g)(x), (h-g)(x), (h g)(x),

and ( h g

)(x).

(h+g)(x) = (1) + ( x4

x3

+ x2

-1) = x4

x3

+ x2

(h-g)(x) = (1) – ( x4

x3

+ x2

-1) = x4

x3

+ x2

+2

(h g)(x) = (1) (x 4

x3

+ x2

-1) = x4

x3

+ x2

-1

(h g

)(x) = 1 x

4

x3

+ x2

1

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Exercises:

1. If h(x) = 7x+3 and g(x) = 2 x2

+1. Find (f+g)(x)

2. If f(x) = x5

-18 and g(x) = x2

– 6x + 9, what is the vaue of (g-h)(x)?

3. If t(x) = 25 x5

and s(x) = 55 x8

, what is the value of ( t s

)(x)?

4. If v(x) = x3

and w(x) = x2

+ 4, solve (v w)(x)?

5. If f(x) = 4x + 11 and g(x) = 5x + 9, nd (f+g)(x).

6. If f(z) = 7z – 4 and g(z) = z-2, nd (f-g)(x).

7. If f(x) =8 x2

-20 and g(x) =-4, nd( f g

)(x).

8. If f(x) = 2x+2 and g(x) = 9 x2

, what is the value of (f g)(x)?

9. If f(x) = 7 x2

+ 8x -3 and g(x) = 7x, solve for (f g)(x)?

10. If f(x) = 35 x8

– 45x and g(x) = 5x, what is the value of ( f g

)(x).

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Answers to Operations on Functions Exercises:

1. 2 x2

+7x +4

2. x5

x2

+ 6x – 27

3. 5

x 11

x3

4. x5

+ 4 x3

5. 9x +20

6. 6z -2

7. 2x 2

+ 5

8. 18 x3

+ 18 x2

9. 49 x3

+ 56 x2

– 21

10.7 x7

– 9

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