Lesson 1

Lesson 1: Functions as Models
A function happen to be a simple mathemical model or a piece of larger model.
Recall that a functon is just a rule or law f, that expresses the dependency of a
variable y, on another variable x.
Example 1: The graph of a function f is shown below:
(a) Find the values f(1) and f(5).
(b) What is the domain and range of f ? Solution:
(a) We see from the graph that the point (1,3) lies on the graph of f, so the
value of f at 1 is f(1) = 3. While x = 5, the graph lies about 0.7 unit below the
x- axis. Therefore, we estimate that f(5) = -0.7.
(b) Notice that f(x) is dened when 0 x 7, so the domain of f is the
closed interval 0,7. See that f takes on all values from the interval -2 to 4, so
the range of f is
-2y 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.
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. (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 Juice2

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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,
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.
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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.
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:
4

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
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
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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:
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,
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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.
7

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)
10

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
11

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).
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(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
15