Detailed Lesson Plan
In
General Mathematics
I. Objective:
At the end of the lesson, students should be able to
1. Evaluate a function
2. Solve problems involving function
II. Subject Matter
Topic:Evaluation of Functions
References: General Mathematics , pages 22- 28 by Orlando A. Oronce
Instructional Materials:
Laptop PowerPoint
LCD projector Activity envelope
Visual Aids
III. Procedure
Teacher’s Activity Student’s Activity
1. Preliminary Activities
1. Prayer
A student will lead the prayer
Class let’s all stand up, Before we start
someone lead the prayer
2. Greetings
Good morning class
Good morning miss Signapan
kindly arrange your chairs, pick up some
pieces of papers Thank you ma’am.
Okay you may now take your seats.
3. Checking of Attendance
Is everybody present today? Yes ma’am, everybody is present.
B. Review
1. Drill
Class, find the answer to the following
multi operation on mathematical
expression by connecting the expression
in column A to the corresponding answer
in column B.
A B
A B
1. (3)(8) – (14+10) 5
1. (3)(8) – (14+10) 5
2. (2+6)(2-3) 100
2. (2+6)(2-3) 100
3. (2)(3) + 4-5 -8
3. (2)(3) + 4-5 -8
4. (2+6)(2) – (-4)
4. (2+6)(2) – (-4)
5. (2)(3) / 3-1 20
5. (2)(3) / 3-1 20
6.[4+2 + (24/6)]2
6.[4+2 + (24/6)] 2
Very good!
2. Review
Ma’am, we solve multi operation
How do we solve multi operation mathematical expression by following the
mathematical expression? rules of PEMDAS.
The student will state the answer.
And what does PEMDAS stand for? Parenthesis, Exponent, Multiplication,
Division, Addition and Subtraction.
That is correct! We solve multi operation
expression by following the operation
order of Parenthesis, Exponent,
Multiplication, Division, Addition,
Subtraction.
C. Lesson Development
Students will raise their hands.
Class, where do usually hear the word Ma’am, we hear the word substitution in
substitution? basketball and volleyball games.
Ma’am, the player will replaced by another
And how do we substitute in basketball? player.
That’s right!
Now class, we will be having an activity
called “Sub me in”
The teacher will present the PowerPoint
presentation of the activity “Sub me in’
Let
A=1 J = 10 S = 19
B=2 K = 11 T = 20
C=3 L = 12 U = 21
D=4 M = 13 V = 22
E=5 N = 14 W = 23
F=6 O = 15 X = 24
G=7 P = 16 Y = 25
H=8 Q = 17 Z = 26
I=9 R = 18
Find the equivalent value of the given
words by substituting the numerical above
to each letter of the word and adding all
the numerical digits.
For example:
B R A I N Yes ma’am.
2+18+1+9+14 =44
Yes ma’am, we are ready.
Do you get it class?
A student will be called to answer.
Are you ready class?
Who would like to try the first word? M A T H
13+1+20+8 = 42
M A T H
B O O K
Now for the second word. 2+15+15+11 = 43
BOOK
S U M
How about the last word? 19+21+13 = 53
SUM
Yes ma’am.
Very good!
Class , does everyone have the same
answer?
Ma’am, we find the numerical value of the
2. Analysis given words.
What did we do in our activity? Ma’am, we substitute the corresponding
value in each letter.
Correct! And how did we find the value of Ma’am, we perform the indicated operation
the given words? and we add all the value.
And then what’s next?
Very good!
Now lets do this activity ( Entry Card).
Find the value of each expression and
write it in the square.
n2 ( 2n – 11)
let n= 15
n2 ( 2n – 11)
n2 ( 2n – 11)
let n= 12
n2 ( 2n – 11)
Ma’am, we substitute 15 in place of n2 to
get 152 or 225.
how did you find the value of n2 when n is Ma’am we substitute 12 in place of n in 2n
15? – 11 to get 2(12) – 11 or 13.
How about the value of 2n - 11 when n is
12?
Very good
How do you use function notation in
evaluating a function?
The function notation y =f(x) tells you that
y is the function of x. if there is rule
relating y to x, such as y= 3x+1, then you
can also write
f(x) =3x+1
e name of
e function is F(x) is read
other letters as “f of x”
ay be used to and this
me the represents
nctions, the value
pecially g and of the
function at
To find f(x) for a given value
of x is to evaluate the function f by
substituting the input value of x into the
equation. the domain is the set of all x
values that makes sense in the equation.
Example:
x-values function-values f(x)=3x+1
x= 2 f(2) = 3(2) + 1 = 7
x= 3 f(3) = 3(3) + 1 = 10
x= 4 f(4) = 3(4) + 1 = 13
x= 5 f(5) = 3(5) + 1 = 16
up to this point , parenthesis have been a
used to represent multiplication . in the
function notation f(x) , there is different
use of parenthesis and remember that 2 students will be called to answer letters
a. f(x) means ‘ the value of f at x “ i. it a and b.
does not mean “ f times x”
b. letters other than f such as g and h Student 1 will answer
can also be used. f(x) = x + 8
c. f is the name of the function and f(x) is f(4) = 4 + 8
the value of the function at x. = 12
Who will try to solve ? Student 2 will answer
If f(x) = x + 8, evaluate each f(x) = x + 8
a. f(4) f(-2) = -2 + 8
b. f(-2) =6
We evaluate functions f by substituting
the value of x into the equation.
Excellent! There is none
3. Abstraction:
Now class, how do we evaluate functions?
Okay is there any further questions?
4. Application
Now that you already know how to
evaluate functions, we will be having a
group activity. I will divide you into 4
groups. And each group will choose their
leaders and the leader will be the one to
ask further instruction from me.
Each group will be given an activity
envelope where you will find the materials
for the activity and the functions that your
group will going to evaluate. you will be
given 5 minutes to solve and we will have
a surprise instruction process on
presenting your answers.
The group will answer
f(x) = 9 – 6x
f(-1) = 9 – 6(-1)
Activity : = 15
Evaluate each function at the indicated f(x) = 9 – 6x
values of the independent variable and f(1) = 9 – 6(1)
simplify the result.
=3
1. f(x) = 9 – 6x
a. f(-1) The group will answer
b. f(1) 1. to find the f(-4), we let x = -4. Since
-4 is less than 0, we used the first
line of the function.
Thus,
f(x) = x2 + 2
= (-4)2 + 2
2. solve the problem = 18
if f(x)= x2 + 2, if x < 0 and 2. to find f(3) we let x = 3. Since 3 is
5x + 2, if x ≥ 0 greater than 0, we used the second
Find; line of the function.
a. f(-4) Thus,
b. f(3) f(x) = 5x + 2
= 5(3) + 2
= 15 + 2
= 17
The student will answer
IV- Evaluation
1. G(x) = 2x2 – 3 , find g(2)
Class, get ½ sheet of paper and answer. G(2) = 2x2 – 3
= 2(2)2 – 3
a. Evaluate each function at the =5
indicated values of the independent 2. F(x) =√ x−1 , find f(5)
variable and simplify the result. F(5) = √ 5−1
=√ 4
1. G(x) = 2x2 – 3 , find g(2) =2
2. F(x) =√ x−1 , find f(5)
3. h(x) = 2x find f(3) 3. h(x) = 2x , find f(3)
2
3 x +4 h(3) = 23
4. f(x) = x , find f(2) =8
2
3 x +4
4. f(x) = , find f(2)
x
2
3(2) + 4
f(2) =
2
3(4 )+ 4
=
2
12+ 4
=
2
=8
V- Assignment :
Solve :
The function C described by
5
C(F) = (F – 32) gives the Celsius
9
temperature corresponding to the
Fahrenheit temperature F.
a. find the Celsius temprature
equivalent to 14℉ .
b. find the Celsius temprature
equivalent to 68℉ .
Prepared by:
Daiku Marte Signapan, LPT
BSED-Mathematics.
