Md.

Abdul Alim
Asst. Professor
Dept. of Sciences, BUFT
Email: abdulalim@buft.edu.bd

Chapter-1
Function
Let 𝑋 and 𝑌 be two sets. A relation between 𝑋 and 𝑌 is a set of ordered pairs, each of the
form (𝑥, 𝑦) where 𝑥 is a member of 𝑋 and 𝑦 is a member of 𝑌. A function from 𝑋 to 𝑌 is a
relation between 𝑋 and 𝑌 that has the property that any two ordered pairs with the same 𝑥-
value also have the same 𝑦-value and is written as 𝑓: 𝑋 → 𝑌. This is read as “𝑓 is a function
of set 𝑋 into set 𝑌”. The variable 𝑥 is called independent variable and the variable 𝑦 is called
dependent variable. The set 𝑋 is called the domain of the function and the set 𝑌 is called the
co-domain of the function. If the element 𝑥 of a set 𝑋 is related to the element 𝑦 of the set 𝑌
then 𝑦 is called the image of 𝑥 and is written as 𝑦 = 𝑓(𝑥). The set of the elements of the set 𝑌
which appear as the images of the elements of the set 𝑋 is called the range of the function.

Definition: A function is a set of ordered pair (x, y) such that each domain value x is associated with
exactly one range value y.

Ex: 𝑦 = 𝑥 2

Domain: The domain is the set of all possible x-values which will make the function "work", and will
output real y-values.

Range: The range of f is the set of all values that the function takes when x takes values in the
domain.

Finding domain and range of a function

When finding the domain, remember:

 The denominator (bottom) of a fraction cannot be zero

 The number under a square root sign must be positive in this work

When finding the range, remember:

 The range of a function is the spread of possible y-values (minimum y-value to

maximum y-value)
 Substitute different x-values into the expression for y to see what is happening. (Ask
yourself: Is y always positive? Always negative? Or maybe not equal to certain
values?)
 Make sure you look for minimum and maximum values of y.
 Draw a sketch! In math, it's very true that a picture is worth a thousand words.

1
Md. Abdul Alim
Asst. Professor
Dept. of Sciences, BUFT
Email: abdulalim@buft.edu.bd

𝒙−𝟑
Ex-1: Find the domain and range of the function 𝒇(𝒙) = 𝟐𝒙+𝟏
𝑥−3
Solution: Let, 𝑦 = 𝑓(𝑥) = 2𝑥+1 … … … … (1)
1
For 𝑥 = − 2 the function in (1) is undefined.
1
Therefore, 𝐷 = ℝ − {− 2}
𝑥−3
From (1) we get, 𝑦 = 2𝑥+1 ⟹ 2𝑥𝑦 + 𝑦 = 𝑥 − 3 ⟹ 2𝑥𝑦 − 𝑥 = −𝑦 − 3
⟹ 𝑥(1 − 2𝑦) = 𝑦 + 3
𝑦+3
⟹𝑥=
1 − 2𝑦
1
Now for 𝑦 = 2 the value of x is undefined.
1
Therefore, 𝑅 = ℝ − {2}

Ex-2: Find the domain and range of the function 𝒇(𝒙) = √𝒙𝟐 − 𝟒𝒙 + 𝟑
Solution: 𝑓(𝑥) = √𝑥 2 − 4𝑥 + 3 = √(𝑥 − 1)(𝑥 − 3)
The given function is defined for (𝑥 − 1)(𝑥 − 3) ≥ 0
Now (𝑥 − 1)(𝑥 − 3) ≥ 0 ⟹ 𝑥 − 1 ≥ 0 𝑎𝑛𝑑 𝑥 − 3 ≥ 0 … … … … (𝑖)
Or, 𝑥 − 1 ≤ 0 𝑎𝑛𝑑 𝑥 − 3 ≤ 0 … … … … (𝑖𝑖)
(𝑖) ⟹ 𝑥 ≥ 1𝑎𝑛𝑑 𝑥 ≥ 3 ⟹ 𝑥 ≥ 3 ⟹ 𝑥 ∈ [3, ∞)
(𝑖𝑖) ⟹ 𝑥 ≤ 1𝑎𝑛𝑑 𝑥 ≤ 3 ⟹ 𝑥 ≤ 1 ⟹ 𝑥 ∈ (−∞, 1]
Therefore 𝐷 = [3, ∞) ∪ (−∞, 1]

Again all possible domain value of x, 𝑦 ≥ 0

Therefore,𝑅 = [0, ∞)
Ex-3: Find the domain and range of the function 𝒇(𝒙) = √𝒙𝟐 + 𝒙 − 𝟏𝟐
Solution: Try Yourself

Even function and Odd function: A function 𝑓(𝑥) is said to be even if 𝑓(−𝑥) = 𝑓(𝑥) and
odd if 𝑓(−𝑥) = −𝑓(𝑥)

Ex-4: Check whether the function 𝒇(𝒙) = 𝐥𝐧 (𝒙 + √𝟏 + 𝒙𝟐 ) is an even function or odd function.

Solution: Given that, 𝑓(𝑥) = ln(𝑥 + √1 + 𝑥 2 )

𝑓(−𝑥) = ln (−𝑥 + √1 + 𝑥 2 )

(√1 + 𝑥 2 − 𝑥)(√1 + 𝑥 2 + 𝑥)
= ln(
√1 + 𝑥 2 + 𝑥

1 + 𝑥2 − 𝑥2
= ln( )
√1 + 𝑥 2 + 𝑥

2
Md. Abdul Alim
Asst. Professor
Dept. of Sciences, BUFT
Email: abdulalim@buft.edu.bd

1
= ln( )
√1 + 𝑥 2 + 𝑥
−1
= ln (√1 + 𝑥 2 + 𝑥)

= − ln(𝑥 + √1 + 𝑥 2 )

= −𝑓(𝑥)

Since, 𝑓(−𝑥) = −𝑓(𝑥) the given function is an odd function.

Ex-5: Determine algebraically whether f(x) = 2x3 – 3x2 – 4x + 4 is even, odd, or neither.

Ex-6: Is f(x) = x/(x2−1) Even or Odd or neither?

Ex-7: Determine whether the following functions are even or odd:

𝟐−𝒙
𝒊)𝒇(𝒙) = 𝒙 − 𝐬𝐢𝐧 𝒙 , 𝒊𝒊) 𝒇(𝒙) = √𝟑 + 𝒙 + 𝒙𝟐 − √𝟑 − 𝒙 + 𝒙𝟐 , 𝑯. 𝑾 𝒊𝒊𝒊) 𝒇(𝒙) = 𝐥𝐧 ( ),
𝟐+𝒙

𝟐+𝒙
𝒇(−𝒙) = 𝐥 𝐧 ( ),
𝟐−𝒙
𝟐−𝒙
=ln (𝟐+𝒙) −𝟏

𝟐−𝒙
=- ln ( )
𝟐+𝒙

=-f(x)

|𝒙| − 𝒙𝟐
𝒊𝒗) 𝒇(𝒙) =
𝐬𝐢𝐧 𝒙

Ex-7: If 𝒇: ℝ → ℝ and 𝒈: ℝ → ℝ are two functions defined by, 𝒇(𝒙) = 𝒙𝟑 − 𝟏 and

𝟏
𝒈(𝒙) = (𝒙 + 𝟏)𝟑 then show that 𝒇 and 𝒈 are inverse to each other.
1
Solution: 𝑓(𝑥) = 𝑥 3 − 1 and 𝑔(𝑥) = (𝑥 + 1)3

1 1 3
Now, (𝑓𝑜𝑔)(𝑥) = 𝑓(𝑔(𝑥)) = 𝑓 ((𝑥 + 1)3 ) = {(𝑥 + 1)3 } − 1 = 𝑥 + 1 − 1 = 𝑥

1 1
And, (𝑔𝑜𝑓)(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔(𝑥 3 − 1) = (𝑥 3 − 1 + 1)3 = (𝑥 3 )3 = 𝑥

Since, (𝑓𝑜𝑔)(𝑥) = (𝑔𝑜𝑓)(𝑥), then 𝑓 and 𝑔 are inverse to each other. (Shown).
𝒙−𝟐
Ex-8: Find the inverse of the function, 𝒇(𝒙) =
𝒙−𝟑

3
Md. Abdul Alim
Asst. Professor
Dept. of Sciences, BUFT
Email: abdulalim@buft.edu.bd

Ex-9: If 𝒇(𝒙) = 𝒙𝟐 + 𝟑𝒙 + 𝟏 and 𝒈(𝒙) = 𝟐𝒙 − 𝟑 then find,

(i) (𝒇𝒐𝒈), (𝒈𝒐𝒇), (𝒇𝒐𝒇)

(ii) (𝒇𝒐𝒈)(𝟑), 𝒈𝒐𝒇(−𝟐)

Composite Function And Inverse

5𝑥+1
1. If 𝑓(𝑥) = 2𝑥 + 3 and 𝑔(𝑥) = 2𝑥−3 then determine (𝑔𝑜𝑓)(𝑥) and (𝑓𝑜𝑔)(𝑥). Also find
(𝑔𝑜𝑓)(−1) and (𝑓𝑜𝑔)(1).

2. Let 𝑓(𝑥) = 3𝑥 − 1 and 𝑔(𝑥) = 5 − 2𝑥. Determine

i. (𝑔𝑜𝑓)(𝑥) and (𝑔𝑜𝑓)(0)
ii. (𝑓𝑜𝑔)(𝑥) and (𝑓𝑜𝑔)(0)
iii. Solve (𝑓𝑜𝑔)(𝑥) = (𝑔𝑜𝑓)(𝑥).

1
3. If 𝑓(𝑥) = 𝑥 2 − 3𝑥 and 𝑔(𝑥) = 𝑥 2 then determine (𝑔𝑜𝑓)(𝑥) and (𝑓𝑜𝑔)(5)
4. Let 𝑓(𝑥) = 3𝑥 − 1 . Determine
i. (𝑓𝑜𝑓)(𝑥)
ii. (𝑓𝑜𝑓𝑜𝑓)(𝑥)
iii. (𝑓𝑜𝑓𝑜𝑓𝑜𝑓)(𝑥)
5. Let f and g be the functions

𝑓(𝑥) = √4𝑥 − 3 and 𝑔(𝑥) = 3𝑥 2 − 6𝑥 + 1

Find the value of (𝑔𝑜𝑓)(3) and then find a general formula for the composite function 𝑔𝑜𝑓.
After having found the general formula, verify that this general value gives same value of
(𝑔𝑜𝑓)(3).

6. For each pair of functions, find (𝑓𝑜𝑔)(𝑥)𝑎𝑛𝑑 (𝑔𝑜𝑓)(𝑥).

2𝑥−1
i. 𝑓(𝑥) = 5𝑥
and 𝑔(𝑥) = 𝑥 2
3 3
ii. 𝑓(𝑥) = 𝑥 − 4 and 𝑔(𝑥) = √𝑥 + 4
iii. 𝑓(𝑥) = sin 𝑥 and 𝑔(𝑥) = 𝑥 2 + 2
1
iv. 𝑓(𝑥) = 2𝑥 2 + 3𝑥 − 2 and 𝑔(𝑥) = 𝑥 2

1 x 5 1 𝑥5 1
7. H.W Show that f ( x)  35 2 x  1 =3(2𝑥 − 1)5 and g ( x)   = 5 + are inverses
486 2 2.3 2
of each other.

4
Md. Abdul Alim
Asst. Professor
Dept. of Sciences, BUFT
Email: abdulalim@buft.edu.bd

5𝑥+1 3𝑥+1
8. Prove that 𝑓(𝑥) = 2𝑥−3 and 𝑔(𝑥) = 2𝑥−5 are inverses of each other.
9. Solve (𝑓𝑜𝑔)(𝑥) = (𝑔𝑜𝑓)(𝑥) where 𝑓(𝑥) = 𝑥 2 𝑎𝑛𝑑 𝑔(𝑥) = 𝑥 + 2