GENMATH - REVIEWER
LESSON 1: FUNCTIONS AND ITS PRESENTATIONS
SOLVING PIECEWISE FUNCTIONS IN REAL LIFE
Example 1: A water station charges ₱35 per container for the first 6 containers, and ₱30 for each
container beyond 6. What is the total cost of buying 10 containers of water?
We are going to use the second function because x is greater than 6 which is 10.
Answer: The total cost of buying 10 containers of water is ₱330.
Example 2: A taxi company charges a flat fee of ₱20 per kilometer for the first 2 kilometers, then
₱15 per kilometer after that. How much is the fare for a 5 kilometer ride?
We are going to use the second function because x is greater than 2 which is 5.
Answer: The fare for a 5 kilometer ride is ₱85.
DETERMINING WHETHER THE RELATION REPRESENTS A FUNCTION OR NOT
Five different ways to represent relations:
1. Ordered Pairs
2. Mapping Diagram
3. Table of Values
4. Equation or Rule
5. Graph
Example 1: A store sells rice for ₱55 per kilo.
If a customer buys 5 kilos, it costs ₱275.
If a customer buys 6 kilos, it costs ₱330.
If a customer buys 7 kilos, it costs ₱385.
▸Ordered Pairs:
(₱275) (₱330) (₱385)
▸Table of Values:
No. of kilos (x)
5
6
7
Cost of rice (y)
₱275
₱330
₱385
▸Mapping Diagram:
X
Y
5
➞
275
6
➞
330
7
➞
385
▸Equation/Rule:
y=55x
Where x is the number of kilos of rice and y is the cost of the rice.
when x=5 then y=55(5)=275
when x=6 then y=55(6)=330
when x=7 then y=55(7)=385
▸Graph:
Range={275, 330, 385}
Domain={5, 6, 7}
Vertical Line Test
It is a way of testing to determine if a relation represented in a graph is a function or not. A graph
represents a function if and only if each vertical line intersects the graph only once.
Example:
LESSON 2: EVALUATING FUNCTIONS
Example 1: Evaluate the function f(x) = 3x2 + 4x - 9 when x = 5.
Solution:
f(x) = 3x2 + 4x - 9
f(5) = 3(5)2 + 4(5) -9
= 3(25) + 20 - 9
= 75 + 11
f(5) = 86
Example 2: Evaluate the function v(m) = 8x2 - 2x + 3 when m = x + 4.
Solution:
v(m) = 8x2 - 2x + 3
v(x + 4) = 8(x + 4)2 - 2(x + 4) + 3
= 8(x2 + 8x + 16) - 2x + 8 + 3
= 8x2 + 64x + 128 - 2x + 11
v(x + 4) = 8x2 + 62x + 39
Example 3: If b(x) = x2 + 2, find b(a + c) and b(a) + b(c).
▸ find b(a + c)
b(x) = x2 + 2
b(a + c) = (a + b)2 + 2
= a2 + 2ab + b2 + 2
b(a + b) = a2 + b2 + 2ab + 2
▸ find b(a) + b(c)
b(x) = x2 + 2
b(a) = a2 + 2
b(c) = c2 + 2
b(a) + b(c) = (a2 + 2) + (c2 + 2)
b(a) + b(c) = a2 + c2 + 4
EVALUATING PIECEWISE FUNCTION
Example 1:
▸Find y when x = 6
We are going to use the second function because x is greater than 3 which is 6.
y = 4x2 + 8x + 3
= 4(6)2 + 8(6) + 3
= 4(36) + 48 + 3
= 144 + 51
y = 195
▸Find y when x = 2
We are going to use the first function because x is less than 3 which is 2.
y = 4x2 + 8x + 3
= 4(2)2 + 8(2) + 3
= 4(4) + 16 + 3
= 16 + 19
y = 35
OPERATIONS ON FUNCTIONS
1. Sum: (f + g)(x) = f(x) + g(x)
2. Difference: (f - g)(x) = f(x) - g(x)
3. Product: (f · g)(x) = f(x) · g(x)
4. Quotient: (f / g)(x) = f(x) / g(x) *fraction
Example 1:
f(x) = 3x + 2
a.) (f + g)(x)
b.) (f - g)(x)
c.) (f · g)(x)
Solution:
a.) (f + g)(x) = f(x) + g(x)
= 3x + 2 + 2x2 - 4
= 2x2 + 3x + 2 - 4
f(x) + g(x) = 2x2 + 3x - 2
b.) (f - g)(x) = f(x) - g(x)
= 3x + 2 - (2x2 - 4)
= 3x + 2 - 2x2 + 4
= 2x2 + 3x + 4 + 2
f(x) - g(x) = 2x2 + 3x + 6
g(x) = 2x2 - 4
c.) (f · g)(x) = f(x) · g(x)
= (3x + 2) (2x2 - 4)
f(x) · g(x) = 6x3 + 4x2 - 12x - 8
Example 2:
f(x) = 4x2 - 9
g(x) = 2x - 3
a.) (f + g)(x)
b.) (f - g)(x)
c.) (f · g)(x)
d.) (f / g)(x)
COMPOSITION OF FUNCTIONS
Equation: (f ⚬ g)(x) = f [g(x)]
Example 1:
f(x) = 2x + 1
a.) (f ⚬ g)(x)
b.) (g ⚬ f)(x)
c.) (f ⚬ f)(x)
d.) (g ⚬ g)(x)
Solution:
a.) (f ⚬ g)(x) = f [g(x)]
f [g(x)] = 2x + 1
g(x) = 4x2 + 2
= 2 (4x2 + 2) + 1
= 8x2 + 4 + 1
f [g(x)] = 8x2 + 5
b.) (g ⚬ f)(x) = g [f(x)]
g [f(x)] = 4x2 + 2
= 4 (2x + 1)2 + 2
= 4x2 + 4x + 1 + 2
g [f(x)] = 4x2 + 4x + 3
c.) (f ⚬ f)(x) = f [f(x)]
f [f(x)] = 2x + 1
= 2 (2x + 1) + 1
= 4x + 2 + 1
f [f(x)] = 4x + 3
d.) (g ⚬ g)(x) = g [g(x)]
g [g(x)] = 4x2 + 2
= 4 (4x2 + 2) + 2
= 16x4 + 16x2 + 4 + 2
g [g(x)] = 16x4 + 16x2 + 6
= Review well
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