This document provides instructions on how to evaluate functions by substituting values for variables and simplifying expressions using order of operations. It gives examples of evaluating various functions, such as f(x)=10+6x^2, by substituting the given value for x and simplifying. The key steps are to substitute the input value for the variable, then simplify the resulting expression using rules like PEMDAS. Floor and ceiling functions are also explained. Practice problems are provided to evaluate functions like f(x)=-5+3x for given x values.

1. LESSON 1. 2
EVALUATING FUNCTIONS
LEARNING COMPETENCY
 EVALUATES A FUNCTION (M11GM-IA-2).

2. • Evaluating a function means finding the value of the
function by replacing or substituting its variable with a
given number or expression. For example, the function
f(x)=10 + 6x2 can be evaluated by squaring the input
value, multiplying by 6, and then adding the product
from 10.
• In this lesson, you are expected to evaluate function
while using the concept of PEMDAS rule (Parenthesis,
Exponent, Multiplication, Division, Addition, and
Subtraction), rules for integers, and law of substitution.

3. Steps on how to evaluate a function
1.Substitute or replace the input variable in the formula
with the value provided.
2.Simplify the expression on the right-hand side of the
equation to obtain the result.

4. Example 1: Given g(x) = 5x – 23, find f(4).
Solution:
g(4) = 5(4) – 23 Substitute 4 for x in the function.
g(4) = 20 – 23 Simplify the expression on the right side
g(4) = -3 of the equation by using PEMDAS and
rules for integers.
Note: 5x means “5 times x”.

5. Example 2: Given f(x)=10 + 6⌈x⌉2 — ⌊x⌋, evaluate f at x = -3.
Solution:
f(x)=10 + 6⌈x⌉2 − ⌊x⌋ Substitute -3 for x in the function.
f(x)=10 + 6⌈-3⌉2 − ⌊-3⌋ Simplify the expression on the right
f(x)=10 + 6(-2)2 − (-4) side of the equation by using
f(x)=10 + 6(4) — (-4) PEMDAS and rules for integers.
f(x)=10 + 24 — (-4)
f(x)=38
Notes:
 Floor function, ⌊x⌋, is the greatest integer that is less than or equal to x.
 Ceiling function, ⌈x⌉, is the least integer that is greater than or equal to x.

6. Example 3: Given p(a) =
4𝑎2−6𝑎+5
𝑎3−4
, evaluate the function at a = 2
Solution:
p(2) =
4(2)2−6(2)+5
(2)3−4
Substitute 4 for x in the function.
p(2) =
4(4)−12+5
8−4
Simplify the expression on the right side
p(2) =
16−12+5
4
of the equation by using PEMDAS and
p(2) =
9
4
rules for integers.

7. Unlock the Process
Fill in the missing number/variable represented by (?) in each given function to unlock the process of
evaluating function.
1. Evaluate f(x) = -5 + 3x for x = 4. 2. Given h(x) = x2 – 3x + 5, find h(-2)
f(?) = -5 + 3(4) h(-2) = (?)2 – 3(?) + 5
f(?) = -5 + (?) h(-2) = (?) – (?) + 5
f(?) = (?) h(?) = 15
3. If t(x) = 4𝑥 − 7, find t(3). 4. Evaluate g(x) = 2x3 + x – 7 for x =1
t(3) = 4(? ) − 7 g(1) = 2(1) + (?) – 7
t(?) = (? ) − 7 (?)(1) = (?) + (?) – 7
t(?) = (? ) g(1) = (?)