Lesson 1

Lesson 1: Functions as Models
A function can 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
quantity, y, on another variable quantity x.
Example 1: 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,
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 2.
1

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:
2

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.
3

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)
4

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
5

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).
6

(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
7

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).
8

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
9