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2 - Evaluation and Operation of Functions
Civil Engineering (BSCE 01)
Ateneo de Davao University
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ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
1
EVALUATION OF FUNCTIONS
In the previous modules, we learned what a function is, and we also identified their domain and range. Now we shall
discover how to evaluate functions. Recall that a function is a “mathematical rule” that assigns a unique value of an input to its
corresponding unique output value. Basically, a function expresses the relationship between an input and output, or between an
independent variable and a dependent variable. Suppose we have the function 𝑦 = 𝑓(𝑥). For this function, 𝑥 is our input value
and 𝑦 is our output value.
Evaluation of functions is basically a two-step process that involves substitution and simplification. Since we have to
specify an input value to a function, we shall use this value and substitute it into the function itself. If the resulting expression can
be simplified, then we have to employ algebraic means and express the final answer in its simplest form.
Sample Problems
- Given 𝑓(𝑥) = 5 − 𝑥
2
, evaluate the function at the given values of the independent variable and simplify
the result
a. 𝑓(0) c. 𝑓(−2)
b. 𝑓(√ 5 ) d. 𝑓(𝑡 − 1)
Before we will begin to solve this problem, let us have a short discussion of what is given here. We have the function
𝑓
(
𝑥
)
which follows that the expression is given in terms of the variable 𝑥. The values 0,√ 5 ,−2 and 𝑡 − 1 are our
input values. Since we are asked to evaluate the function, we shall replace all variables in the function with the different
input values.
Solution: 𝑓
(
𝑥
)
= 5 − 𝑥
2
a.)
𝑓( 0 )= 5 −( 0 )
2
𝑓
(
0
)
= 5
b.)
𝑓(√ 5 )= 5 −(√ 5 )
2
𝑓(√ 5 )= 5 − 5
𝑓(√ 5 )= 0
c.)
𝑓
(
−
)
= 5 −
(
−
)
2
𝑓
(
−
)
= 5 −
(
4
)
𝑓(−2)= 1
ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
2
d.)
𝑓
(
𝑡 − 1
)
= 5 −
(
𝑡 − 1
)
2
𝑓
(
𝑡 − 1
)
= 5 −
(
𝑡
2
− 2𝑡 + 1
)
𝑓
(
𝑡 − 1
)
= 5 − 𝑡
2
+ 2𝑡 − 1
𝑓(𝑡 − 1)= 4 − 𝑡
2
+ 2𝑡
Notice that for the first three (3) items, our input values are constants. Therefore, we expect that our output value is also
a constant. In the last item, our input value consists of the variable 𝑡. It follows that our final answer is still a function,
but this time, in terms of the new variable. As previously stated, our input value, or independent variable, can take on
many forms.
- If 𝑓(𝑥)= 3𝑥 − 1, evaluate the expression
𝑓
(
𝑥
)
− 𝑓(1)
𝑥 − 1
In this problem, we are asked to evaluate the expression given above. Notice that it involves 𝑓(𝑥) and 𝑓(1). We are
already given 𝑓(𝑥) so all that is left for us is to evaluate 𝑓(1).
Solution:
𝑓
(
1
)
= 3
(
1
)
− 1
𝑓( 1 )= 2
Now let us substitute the data that we obtained into the expression
𝑓
(
𝑥
)
− 𝑓
(
1
)
𝑥 − 1
=
3𝑥 − 1 − 2
𝑥 − 1
=
3𝑥 − 3
𝑥 − 1
The expression we just obtained above can still be simplified by factoring out 3 from the numerator. We see, then, that
we can cancel out common terms.
=
3(𝑥 − 1)
𝑥 − 1
𝑓
(
𝑥
)
− 𝑓(1)
𝑥 − 1
= 3
ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
4
OPERATIONS INVOLVING FUNCTIONS
After learning how to evaluate functions, this time we will tackle the different operations involving functions. Note that
operation here refers to the basic arithmetic operations such as addition, subtraction, multiplication and division. The goal of this
section is to combine two or more functions by means of arithmetic operations. Often times, a problem may specify two or more
functions and they express their relationship using arithmetic operations. When performing basic operations with functions, the
result may yield either a new function, a constant value or the function itself. At the end of this section, we shall also tackle a
special operation for functions which is the composition of functions.
Basic Operations Involving Functions:
Suppose 𝑓(𝑥) and 𝑔(𝑥) are two different functions involving the same independent variable, 𝑥
- Sum Rule
(
𝑓 + 𝑔
)(
𝑥
)
= 𝑓
(
𝑥
)
+ 𝑔(𝑥)
Difference Rule (𝑓 − 𝑔)(𝑥)= 𝑓(𝑥)− 𝑔(𝑥)
Product Rule
(
𝑓 ∙ 𝑔
)(
𝑥
)
= 𝑓
(
𝑥
)
∙ 𝑔(𝑥)
- Quotient Rule (
𝑓
𝑔
)
(
𝑥
)
=
𝑓(𝑥)
𝑔(𝑥)
Sample Problems
- Three functions are defined by 𝑓
(
𝑥
)
= 𝑥 + 1,𝑔
(
𝑥
)
=√3𝑥,ℎ
(
𝑥
)
=
1
6−𝑥
. Solve for (
𝑔
ℎ
)(3) and
(𝑓 ∙ ℎ)(−2)
Solution:
a.)
The notation (
𝑔
ℎ
)( 3 ) can be understood in two ways. First, we can start by getting the quotient by dividing 𝑔(𝑥) to ℎ(𝑥)
and then evaluating the result by substituting 3 into the resulting function. An alternative way for this is to use the formula
(
𝑔
ℎ
)
(
3
)
=
𝑔
( 3
)
ℎ
( 3
)
. Either way, we will be able to obtain the same answer. For this problem we will apply the first method and
then apply the second method in the next item so that you can decide for yourself which method best suits you.
𝑔(𝑥)
ℎ(𝑥)
=
√
3𝑥
1
6 − 𝑥
=
√
3𝑥(6 − 𝑥)
Evaluating at 𝑥 = 3
=√ 3
( 3
) (6 − 3)
=√ 9 (3)
= 3(3)
(
𝑔
ℎ
)( 3 )= 9
ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
5
b.)
The notation (𝑓 ∙ ℎ)(−2), using the second method, can be rewritten as 𝑓(−2) ∙ ℎ(−2). That is, to evaluate the two
functions separately at the indicated input value and then multiplying the result thereafter.
𝑓
(
−
)
= −2 + 1
𝑓
(
−
)
= −
ℎ
(
−
)
=
1
6 − (−2)
ℎ
(
−
)
=
1
6 + 2
ℎ
(
−
)
=
1
8
(𝑓 ∙ ℎ)(−2)= −1(
1
8
)
(
𝑓 ∙ ℎ
)(
−
)
= −
1
8
- If 𝑓
(
𝑥
)
=
3𝑥−
𝑥
2
−
and 𝑔
(
𝑥
)
=
2𝑥+
𝑥
2
−
solve for
(
𝑓 + 𝑔
)(
𝑥
)
,
(
𝑓𝑔
)(
𝑥
)
and (
𝑓
𝑔
)(𝑥)
Solution:
For this certain problem, we are given two functions which are both rational expressions. What is asked of us is to solve the
resulting function if we add, multiply and divide the two. So we will be performing three (3) sets of solutions here.
a.)
(
𝑓 + 𝑔
)(
𝑥
)
= 𝑓
(
𝑥
)
+ 𝑔(𝑥)
=
3𝑥 − 1
𝑥
2
− 25
+
2𝑥 + 4
𝑥
2
− 25
Recall from basic algebra that when adding rational expressions with the same denominator, we can simply perform
addition/subtraction (depending on the operation involved) to combine like terms in the numerator.
=
3𝑥 − 1 + 2𝑥 + 4
𝑥
2
− 25
=
5𝑥 + 3
𝑥
2
− 25
ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
7
COMPOSITION OF FUNCTIONS
Composition of functions is a special type of operation on functions. Similar to the previous topic, it still involves two
or more functions. But for composition of functions, we are dealing with a function within a function. Consider the definition
below.
To solve a simple composite function problem, say 𝑓(𝑔
(
𝑥
)
), take the function 𝑔(𝑥) and then substitute the entire
function into 𝑓(𝑥). If possible, try to simplify the answer. A composite function may have more than two functions within functions.
Suppose (𝑓 𝑜 𝑔 𝑜 ℎ)(𝑥), this is considered to be equal to 𝑓(𝑔(ℎ
(
𝑥
)
)). To solve problems involving composite functions of
three or more functions, always begin on the inner composite function, and progressing to the outer composite function.
Sample Problem
- If 𝑓
(
𝑥
)
=
3
𝑥
and 𝑔
(
𝑥
)
= 𝑥
2
− 1, solve for (𝑔 𝑜 𝑓)(𝑥) and (𝑓 𝑜 𝑔)(𝑥)
Solution:
a.)
(
𝑔 𝑜 𝑓
)(
𝑥
)
= 𝑔(𝑓
(
𝑥
)
)
The expression above suggests that we take the function 𝑓(𝑥) first and then substitute it into 𝑔(𝑥). Therefore, we should
be able to get
=
(
3
𝑥̇
)
2
− 1
=
9
𝑥
2
− 1
=
9 − 𝑥
2
𝑥
2
b.)
(
𝑓 𝑜 𝑔
)(
𝑥
)
= 𝑓(𝑔
(
𝑥
)
)
The expression above suggests that we take the function 𝑔(𝑥) first and then substitute it into 𝑓(𝑥). Therefore, we should
be able to get
Definition of a Composite Function
Let f and g be two functions of the same independent variable, say 𝑥.
The function
(
𝑓 𝑜 𝑔
)(
𝑥
)
= 𝑓(𝑔
(
𝑥
)
) is the composite function of 𝑓
with 𝑔. The domain of 𝑓 𝑜 𝑔 is the set of all 𝑥 in the domain of 𝑔
such that 𝑔(𝑥) is in the domain of 𝑓.
ESM 1030 – ENGINEERING CALCULUS 1 (DIFFERENTIAL CALCULUS)
EVALUATION OF FUNCTIONS AND OPERATIONS INVOLVING FUNCTIONS
Prepared by: Engr. Nelson John M. Namuag
Engineering Sciences and Mathematics Department
8
=
3
𝑥
2
− 1
PRACTICE PROBLEMS
- If 𝑓(𝑥)= {
(
−𝑥
)
2
3 𝑓𝑜𝑟 𝑥 < 0
4 𝑓𝑜𝑟 𝑥 = 0
𝑥 + 2 𝑓𝑜𝑟 𝑥 > 0
solve for 𝑓(−8),𝑓( 0 ),𝑓(1)
- For 𝑓
(
𝑥
)
= 𝑥
2
- 4𝑥, find
𝑓
( 𝑥+ℎ
) −𝑓(𝑥)
ℎ
- Given the function ℎ(𝑥)=
3𝑥
2
−
𝑥
2
, evaluate ℎ(−
2
3
) and ℎ(𝑎 − 2)
- If 𝑓
(
𝑥
)
=
𝑥+
𝑥−
and 𝑔
(
𝑥
)
=
1
𝑥
, find
(
𝑓 + 𝑔
)(
𝑥
)
and (𝑓𝑔)(𝑥)
- Compute for (𝑓 𝑜 𝑔)(𝑐) for 𝑓
(
𝑥
)
= √𝑥
2
− 36, 𝑔
(
𝑥
)
= 𝑥
2
− 3𝑥 if 𝑐 = 5
Answers
𝑓
( −
) = 4, 𝑓
( 0
) = 4, 𝑓
( 1
) = 3
𝑓
( 𝑥 + ℎ
) − 𝑓(𝑥)
ℎ
= 2𝑥 + 4 + ℎ
ℎ (−
2
3
) = −
33
4
, ℎ
( 𝑎 − 2
) =
3𝑎
2
− 12𝑎 + 7
(𝑎 − 2)
2
(𝑓 + 𝑔)(𝑥)=
𝑥
2
- 2𝑥 − 1
𝑥(𝑥 − 1)
,(𝑓𝑔)(𝑥)=
𝑥 + 1
𝑥(𝑥 − 1)
𝑓(𝑔
( 𝑐
) ) = 8
2 - Evaluation and Operation of Functions
Course: Civil Engineering (BSCE 01)
University: Ateneo de Davao University
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