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. 

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)] 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

g o f 

Illustration :

Consider the set A of all students, who appeared in class X of Board Examination in March 2020. 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. 

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.

Problem 1 :

Find f o g and g o f when f(x) = 2x + 1 and g(x) = x² - 2.

Solution :

f o g :

= f[g(x)]

= f[x² - 2]

= 2(x² - 2) + 1

= 2x² - 4 + 1

= 2x² - 3 ----(1)

g o f :

= g[f(x)]

= g[2x + 1]

= (2x + 1)² - 2

= (2x)² + 2(2x)(1) + 1² - 2

= 4x² 4x + 1 - 2

= 4x² 4x - 1 ----(2)

From (1) and (2), we see that f o g ≠ g o f.

Problem 2 :

Represent the following function as a composition of two functions.

h(x) = √(2x² - 5x + 3)

Solution :

Let g(x) = 2x² - 5x + 3 and f(x) = √x.

h(x) = √g(x)

= f[g(x)]

= f o g(x)

Problem 3 :

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

Solution :

f o g = g o f

f[g(x)] = g[f(x)]

f[2x + k] = g[3x - 2]

3(2x + k) - 2 = 2(3x - 2) + k

6x + 3k - 2 = 6x - 4 + k

Subtract 6x from each side. 

3k - 2 = -4 + k

Subtract k from each side. 

2k - 2 = -4

Add 2 to each side. 

2k = -2

Divide each side by 2.

k = -1

Problem 4 :

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

Solution :

f o f(k) = 5

f[f(k)] = 5

f[2k - 1] = 5

2(2k - 1) - 1 = 5

4k - 2 - 1 = 5

4k - 3 = 5

Add 3 to each side.  

4k = 8

Divide each side by 2.

k = 2

Composition of Three Functions

Problem 5 :

f(x) = 2x + 3 , g(x) = 1 - 2x and h(x) = 3x . Prove that

f o (g o h) = (f o g) o h

Solution :

f o (g o h) :

(f o g) o h :

From (1) and (2), 

f o (g o h) = (f o g) o h

Problem 6 :

If f(x) = 3x + 1, g(x) = x + 3 and gff(x) = fgg(x), find x. 

Solution :

gff(x) = fgg(x)

g{f[f(x)]} = f{g[g(x)]}

g{f[3x + 1]} = f{g[x + 3]}

g{3(3x + 1) + 1} = f{(x + 3) + 3}

g{9x + 3 + 1} = f{x + 6}

g{9x + 4} = f{x + 6}

(9x + 4) + 3 = 3(x + 6) + 1

9x + 4 + 3 = 3x + 18 + 1

9x + 7 = 3x + 19

Subtract 3x and 7 from each side. 

6x = 12

Divide each side by 6. 

x = 2

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