Let f : A → B and g : B → C be two functions (Fig). Then the composition of f and g denoted by g o f is defined as the function g o f (x ) = g( f (x )) for all x ∈ A .

**Composition of Functions**

When a car driver
depresses the accelerator pedal, it controls the flow of fuel which in turn influences
the speed of the car. Likewise, the
composition of two functions is a
kind of ‘chain reaction’, where the functions act upon one after another
(Fig.1.40).

We can explain this
further with the concept that a function is a ‘process’. If *f *and *g *are
two functions then the composition *g*(*f *(*x*)) (Fig.1.41) is
formed in two steps.

(i) Feed an input (say
*x*) to *f *;

(ii) Feed the output *f*(*x*)
to *g *to get *g*(*f *(*x*)) and call it *gf*(*x*).

Consider the set *A *of
all students, who appeared in class *X *of Board Examination. Each student
appearing in the Board Examination is assigned a roll number. In order to have confidentiality, the Board
arranges to deface the roll number of each student and assigns a code number to
each roll number.

Let *A *be the
set of all students appearing for the board exam. *B *⊆ **N** be the set all roll numbers and* C *⊆ **N** be the set of all code numbers (Fig.1.41). This gives rise to
two functions *f*: *A *→ *B* and *g *: *B *→ *C *given
by *b *= *f *(*a*) be the roll number assigned to student *a*,
*c *= *g*(*b*) be the code number assigned to roll number *b*,
where *a *∈ *A *, *b *∈ *B* and *c *∈ *C.*

We can write *c *=
*g*(*b*) = *g*(*f *(*a*)).

Thus, by the
combination of these two functions, each student is eventually attached a code
number. This idea leads to the following definition.

Let *f* : *A*
→ *B* and *g* : *B* → *C* be two functions (Fig.1.42).
Then the composition of *f* and *g* denoted by *g o f* is defined as the function *g o f*
(*x* ) = *g*( *f*
(*x* )) for all *x* ∈ *A* .

**Example 1.20**

Find *f *o *g *and *g *o *f* when *f *(*x*) = 2*x *+
1 and *g*(*x*) = *x*^{2} – 2

*Solution*

* **f *(*x*) = 2*x *+
1 , *g*(*x*) = *x*^{2} – 2

*f *o *g*(*x*) = *f *(*g*(*x*))
= *f *(*x*^{2} − 2) = 2(*x*^{2} − 2) + 1 = 2*x*^{2}
– 3

*g *o *f *(*x*) = *g*(*f *(*x*))
= *g*(2*x *+ 1) = (2*x *+ 1)^{2} − 2 = 4*x*^{2}
+ 4*x *– 1

Thus *f *o *g *= 2*x*^{2} − 3, *g *o *f *= 4*x*^{2} + 4*x *−
1. From the above, we see that *f *o *g *≠ *g *o *f .*

**Note**

Generally,** ***f o g*** ≠ ***g o f***
**for any two functions** ***f*** **and** ***g*. So, composition of functions is**
**not commutative.

**Example 1.21**

Represent the function
*f *(*x*) = as a composition of two functions.

*Solution*

We set* f *(*x*)
= 2*x*^{2} − 5*x *+ 3 and *f *(*x*) = √*x*

Then,

**Example 1.22**

If *f *(*x*)
= 3*x *− 2 , *g*(*x*) = 2*x *+ *k* and if *f *o *g *= *g *o *f *, then find
the value of *k*.

*f *(*x*) = 3*x *−
2 , *g*(*x*) = 2*x *+ *k*

*f *o *g*(*x*) = *f *(*g*(*x*))
= *f *(2*x *+ *k*) = 3(2*x *+ *k*) − 2 = 6*x *+ 3*k
*– 2

Thus, *f *o *g*(*x*) = 6*x *+ 3*k *–
2.

*g *o *f *(*x*) = *g*(3*x *− 2)
= 2(3*x *− 2) + *k*

Thus, *g *o *f *(*x*) = 6*x *− 4 + *k *.

Given that *f *o *g *= *g *o *f*

Therefore, 6*x *+
3*k *− 2 = 6*x *− 4 + *k*

6*x *− 6*x *+
3*k *− *k *= −4 + 2 ⇒ *k *= −1

Find *k *if *f *o *f *(*k*) = 5 where *f *(*k*)
= 2*k *– 1.

*f *o *f *(*k*) = *f *(*f *(*k*))

= 2(2*k *− 1) − 1
= 4*k *− 3

Thus, *f *o *f *(*k*) = 4*k *– 3

But, it is given that *f
*o *f *(*k*) =
5

Therefore 4*k *-
3 = 5 ⇒ *k *= 2 .

Let *A*,
*B*, *C*, *D *be four sets and let *f *: *A *→ *B *,
*g *: *B *→ *C*and *h *: *C *→ *D
*be three functions (Fig.1.43). Using composite functions *f *o *g *and *g *o *h *, we get two new functions
like (*f *o *g*) o *h *and *f *o (*g *o *h*).

We observed that the
composition of functions is not commutative. The natural question is about the
associativity of the operation.

Composition of three
functions is always associative. That is, f o (g o h) = (f o g) o h

If *f *(*x*)
= 2*x *+ 3 ,* g*(*x*) = 1 − 2*x and h*(*x*) = 3*x *.
Prove that* f *o (*g *o *h*) = (*f *o *g*) o *h*

*f *(*x*) = 2*x *+
3 , *g*(*x*) = 1 − 2*x *, *h*(*x*) = 3*x*

Now, (*f *o *g*)(*x*) = *f *(*g*(*x*))
= *f *(1 − 2*x*) = 2(1 − 2*x*) + 3 = 5 − 4*x*

Then, (*f *o *g*) o *h*(*x*) = (*f *o *g*)(*h*(*x*)) = (*f *o *g*)(3*x*) = 5 − 4(3*x*) = 5 −
12*x ……… *(1)

(*g *o *h*)(*x*) = *g*(*h*(*x*))
= *g*(3*x*) = 1 − 2(3*x*) = 1 − 6*x*

So, *f *o (*g *o *h*)(*x*) = *f *(1 − 6*x*) = 2(1 − 6*x *)
+ 3= 5 − 12*x ………*(2)

From (1) and (2), we
get (*f *o *g*) o *h *= *f *o (*g *o *h*)

Find *x *if *gff*(*x*)
= *fgg*(*x*), given *f *(*x*) = 3*x *+ 1 and *g*(*x*)
= *x *+ 3 .

*gff*(*x*) = *g *[*f
*{*f *(*x*)}] (This means “*g *of *f *of *f *of *x*”)

= *g *[ *f *(3*x
*+1)] = *g *[ 3(3*x *+1)+1] = *g *(9*x *+ 4)

*g *(9*x *+ 4) = [ (9*x
*+ 4) + 3] = 9*x *+ 7

*fgg*(*x*) = *f *[*g
*{*g *(*x*)}] (This means “*f *of *g *of *g *of *x*”)

= *f *[ *g *(*x
*+ 3)] = *f *[ (*x *+ 3) + 3] = *f *(*x *+ 6)

*f *(*x *+ 6) = [ 3(*x
*+ 6) + 1 ] = 3*x *+ 19

These two quantities
being equal, we get 9*x *+ 7 = 3*x *+ 19. Solving this equation we
obtain *x *= 2.

Tags : Definition, Illustration, Example, Solution | Mathematics , 10th Mathematics : UNIT 1 : Relation and Function

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10th Mathematics : UNIT 1 : Relation and Function : Composition of Functions | Definition, Illustration, Example, Solution | Mathematics

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