Composite Functions
Let f : A->B and g : B->C be two functions. Then the composition of f and g, denoted by g o f, is defined as the function g o f : A->C given by g o f (x) = g{f(x)}, ∀ x ∈ A.

Clearly, dom(g o f) = dom(f).
Also, g o f is defined only when range(f) is a subset of dom(g).
Table of Contents
Evaluating composite functions
We know composite function is written as f o g(x), g o f(x) and so on. Here f o g(x) will be evaluated as f{g(x)} and g o f(x) will be evaluated as g{f(x)}.
Solved Examples
Problem 1: Let f : {2, 3, 4, 5} -> {3, 4, 5, 9} and g : {3, 4, 5, 9} -> {7, 11, 15} be functions defined as
f(2) = 3, f(3) = 4, f(4) = f(5) = 5 and g (3) = g(4) = 7 and g(5) = g(9) = 11. Find g o f(x).
Solution:
g o f(x) = g{ f(x)}. So first we find the inner bracket which is f(x) here.
We have the values of f(2), f(3), f(4) and f(5), hence we have to find the values of g o f(x) for all these values.
g o f(2) = g{ f(2) } = g(3) = 7,
Best JEE & NEET Physics, Chemistry, Biology, and Math Study Material – Anand Classes by Neeraj Anand (Published by Anand Technical Publishers)
If you are preparing for JEE Mains, JEE Advanced, or NEET, having the right study material can make all the difference. Anand Classes, authored by Neeraj Anand and published by Anand Technical Publishers, provides one of the most comprehensive, structured, and exam-oriented study materials for Physics, Chemistry, Biology, and Mathematics Subjects according to latest trends of JEE & NEET Entrance Exams.
Why Choose Anand Classes Study Material for JEE & NEET Preparation?
The study materials by Neeraj Anand are designed to simplify complex concepts, provide in-depth explanations, and offer ample practice questions to help students achieve top scores in competitive exams.
✅ Comprehensive Coverage – Detailed explanations of Physics, Chemistry, Biology, and Mathematics concepts.
✅ JEE & NEET Focused – Designed specifically for competitive exam success.
✅ Solved Examples & Practice Questions – Strengthen your understanding with concept-based problems.
✅ Short Tricks & Formulas – Easy-to-remember techniques for quick problem-solving.
✅ Concept Clarity: Easy-to-understand theory with step-by-step explanations.
✅ Topic-Wise Breakdown: Well-structured chapters following the latest syllabus of JEE Mains, JEE Advanced, and NEET.
✅ Solved Examples & Practice Questions: Includes previous year questions (PYQs), important formulas, and shortcut techniques.
✅ NCERT-Based & Advanced Level: Covers both board exam preparation and entrance exam syllabus for JEE & NEET aspirants.
✅ Time-Saving Tricks: Quick formulas, memory techniques, and problem-solving strategies.📚 Subjects Covered in Anand Classes Study Material
🔵 Physics (For JEE & NEET)
- Kinematics & Laws of Motion
- Work, Power & Energy
- Gravitation & Fluid Mechanics
- Thermodynamics & Heat Transfer
- Electrostatics & Magnetism
- Optics & Modern Physics
- Semiconductor Electronics & Communication
🟠 Chemistry (For JEE & NEET)
Physical Chemistry
- Mole Concept & Stoichiometry
- Thermodynamics & Chemical Equilibrium
- Electrochemistry & Chemical Kinetics
Inorganic Chemistry
- Periodic Table & Chemical Bonding
- Coordination Compounds & Metallurgy
- P-Block, D-Block & F-Block Elements
Organic Chemistry
- Hydrocarbons & Functional Groups
- Reaction Mechanisms & Named Reactions
- Biomolecules & Polymers
🟢 Biology (For NEET)
- Diversity in the Living World
- Cell Structure & Function
- Genetics & Evolution
- Human Physiology & Reproduction
- Ecology & Environment
🔴 Mathematics (For JEE Mains & Advanced)
Permutation, Combination & Complex Numbers
Algebra & Probability
Trigonometry & Coordinate Geometry
Calculus (Differential & Integral)
Vectors & 3D Geometry
📥 Download PDF & Purchase the Study Material
The Anand Classes Study Material is available in PDF format and hardcopy. You can access high-quality notes, question banks, and practice tests to boost your JEE & NEET preparation.
🔹 Download Now: Get the latest edition PDF for easy access on mobile, tablet, or laptop.
🔹 Buy Hardcopy: Order the printed book for detailed study and offline preparation.📌 Published by: Anand Technical Publishers
📌 Author: Neeraj Anand
g o f (3) = g{ f(3) } = g(4) = 7,
g o f(4) = g{ f(4) } = g(5) = 11
And g o f(5) = g(5) = 11.
Problem 2: Show that if f : A -> B and g : B -> C are onto, then g o f : A -> C is also onto.
Solution
Given an arbitrary element z ∈ C, there exists a pre-image y of z under g such that g(y) = z, since g is onto.
Further, for y ∈ B, there exists an element x in A with f(x) = y, since f is onto.
Therefore, g o f(x) = g{ f(x) } = g(y) = z, showing that g o f is onto.
Problem 3: Let ƒ: R->R:f(x) = (x2 – 3x + 2). Find f o f(x).
Solution:
f o f(x) = f {f(x) } = f(x2 – 3x + 2) = f(y) (let y = x2 – 3x + 2)
= y2 – 3y + 2
= (x2 – 3x + 2) – 3(x2 – 3x + 2) + 2
= x4 – 6x3 + 10x2 – 3x.
Evaluating Composite Functions: Using Tables
In this type of questions, we will be given a table with values of x, f(x), g(x) and we will need to find the composite of f(x) and g(x) like in example 1 we are asked to find f o g(1).
To find the solution, we will start from the inner bracket and find its value in the table given, so we will find the value of g(1) from the table which is 4. Then for the outer bracket again we will find its value from the given table, so we will find the value of f(4) from the table which is 2. Hence, we will get the required answer which is 2 in this case.
Problem 1: Using the table below, evaluate f o g(1) and g o f(4).
x | f(x) | g(x) |
---|---|---|
1 | 5 | 4 |
2 | 8 | 5 |
3 | 4 | 3 |
4 | 2 | 8 |
Solution:
So, f o g(1) = f{g(1)} and g o f(4) = g{f(4)}.
We first evaluate the inner bracket then the outer bracket, using the values given in table.
For f{g(1)}, g(1) = 4. Now f(4) = 2.
Hence, f{g(1)} = 2.
Similarly, for g{f(4)}, f(4) = 2. Now g(2) = 8.
Hence, g{f(4)} = 8.
Therefore, f o g(1) = f{g(1)} = f(4) = 2 and g o f(4) = g{f(4)} = g(2) = 8.
Problem 2: Using the table below, evaluate f o g(3) and f o f(1).
x | f(x) | g(x) |
1 | 3 | 2 |
2 | 7 | 3 |
3 | 8 | 5 |
4 | 6 | 7 |
5 | 2 | 9 |
Solution:
So, f o g(3) = f{g(3)} and f o f(1) = f{f(1)}.
We first evaluate the inner bracket then the outer bracket, using the values given in table.
For f{g(3)}, g(3) = 5. Now f(5) = 2.
Hence, f{g(3)} = 2.
Similarly, for f{f(1)}, f(1) = 3. Now f(3) = 8.
Hence, f{f(1)} = 8.
Therefore, f o g(3) = f{g(3)} = f(5) = 2 and f o f(1) = f{f(1)} = f(3) = 8.
Evaluating composite functions: Using Graphs
In this type of questions, we will be given a graph having f(x) and g(x) curve, and we will be asked to find a composite of f(x) and g(x) like in example 1 we have to find f o g(2).
Again we will start from the inner bracket, so we have to find g(2). So we will see the g(x) curve. So at x=2, we find that y=3. Now we will need to find f(3). So we will see the f(x) curve. Now we will look at x=3, there we find y=4. Hence, we get the required solution which is 3 in example 1.
Problem 1: Using the graphs below, evaluate f o g(2).


Solution:
So, f o g(2) = f{ g(2) }
From the graph of g(x), we need to find g(2),
when x = 2, y = 3, hence g(2) = 3
Now f o g(2) = f(g(2)) = f(3).
From the graph of f(x), we need to find f(3),
when x = 3, y = 4, hence f(3) = 4
Therefore, f o g(2) = f{g(2)} = f(3) = 4.
Problem 2: Using the graph below, evaluate g o h(5).

Solution:
So, g o h(5) = g{ h(5) }
From the graph of h(x), we need to find h(5),
When x = 5, y = -2, hence h(5) = -2
Now g o h(5) = g{h(5)} = g(-2).
From the graph of g(x), we need to find g(-2),
when x = -2, y = 3, hence g(-2) = 3
Therefore, g o h(5) = g{h(5)} = g(-2) = 3.