Composite functions-Relations & functions Class 12 Math Chapter 1 Notes Study Material Download free pdf

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.

Composite-Functions

Clearly, dom(g o f) = dom(f).

Also, g o f is defined only when range(f) is a subset of dom(g).

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,

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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) = (x– 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).

xf(x)g(x)
154
285
343
428

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).

xf(x)g(x)
132
273
385
467
529

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).

Composite-Functions-2
Composite-Functions-3

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).

Composite-Functions-4

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.

Er. Neeraj K.Anand is a freelance mentor and writer who specializes in Engineering & Science subjects. Neeraj Anand received a B.Tech degree in Electronics and Communication Engineering from N.I.T Warangal & M.Tech Post Graduation from IETE, New Delhi. He has over 30 years of teaching experience and serves as the Head of Department of ANAND CLASSES. He concentrated all his energy and experiences in academics and subsequently grew up as one of the best mentors in the country for students aspiring for success in competitive examinations. In parallel, he started a Technical Publication "ANAND TECHNICAL PUBLISHERS" in 2002 and Educational Newspaper "NATIONAL EDUCATION NEWS" in 2014 at Jalandhar. Now he is a Director of leading publication "ANAND TECHNICAL PUBLISHERS", "ANAND CLASSES" and "NATIONAL EDUCATION NEWS". He has published more than hundred books in the field of Physics, Mathematics, Computers and Information Technology. Besides this he has written many books to help students prepare for IIT-JEE and AIPMT entrance exams. He is an executive member of the IEEE (Institute of Electrical & Electronics Engineers. USA) and honorary member of many Indian scientific societies such as Institution of Electronics & Telecommunication Engineers, Aeronautical Society of India, Bioinformatics Institute of India, Institution of Engineers. He has got award from American Biographical Institute Board of International Research in the year 2005.