HOW TO FIND COMPOSITION OF TWO 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 (x) = g(f (x)) for all x ∈ A.

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

Using the functions f and g given, find f o g and g o f. Check whether  f o g = g o f .

Question 1 : 

f(x) = x - 6 and g(x) = x²

Solution : 

f o g = f[g(x)]

= f[x²]

= x² - 6 ----(1)

g o f = g[f(x)]

=  g[x- 6]

= (x - 6)²

= x² - 2(x)(6) + 6²

= x² - 12x + 36 ----(2)

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

Question 2 : 

f(x) = 2/x and g(x) = 2x² - 1

Solution : 

f o g = f[g(x)]

= f[2x² - 1]

= f(2x² - 1)

= 2/(2x² - 1) ----(1)

g o f = g[f(x)]

= g[2/x]

 = 2(2/x)² - 1

= 2(4/x²) - 1

= 8/x² - 1 ----(2)

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

Question 3 :

f(x) = (x + 6)/3 and g(x) = 3 - x 

Solution : 

f o g = f[g(x)]

= f[3 - x]

= (3 - x + 6)/3

= (9 - x)/3 ----(1)

g o f = g[f(x)]

= g[(x + 6)/3]

= 3 - [(x + 6)/3]

= 9/3 - [(x + 6)/3]

=  [9 - (x + 6)]/3

= (9 - x - 6)/3

= (3 - x)/3 ----(2)

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

Question 4 :

f(x) = 3 + x and g(x) = x - 4

Solution : 

f o g = f[g(x)]

= f[x - 4]

= 3 + x - 4

= x - 1 ----(1)

g o f = g[f(x)]

= g[3 + x]

= 3 + x - 4 

= x - 1 ----(2)

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

Question 5 :

f(x) = 4x² - 1 and g(x) = 1 + x

Solution :

f o g = f[g(x)]

= f[1 + x]

= 4(1 + x)² - 1

= 4(1 + 2x + x²) - 1

= 4 + 8x + 4x² - 1

= 4x² + 8x + 3 ----(1)

g o f = g[f(x)]

= g[4x² - 1]

= 1 + 4x² - 1

 = 4x² ---(2)

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

Question 6 :

Let f(x) = 1 - 2x and h(x) = f(g(x)). Fill in the table.

Solution :

f(x) = 1 - 2x

h(x) = f(g(x))

When x = -3

h(-3) = f(g(-3))

= f(-2)

Evaluating f(-2) using the function 1 - 2x :

f(-2) = 1 - 2(-2)

= 1 + 4

= 5

When x = -1

h(-1) = f(g(-1))

= f(1)

Evaluating f(1) using the function 1 - 2x :

f(1) = 1 - 2(1)

= 1 - 2

= -1

When x = 0

h(0) = f(g(0))

= f(2)

Evaluating f(2) using the function 1 - 2x :

f(2) = 1 - 2(2)

= 1 - 4

= -3

Question 7 :

Let g(x) = 3^x and h(x) = g(f(x)). Fill in the table.

Solution :

g(x) = 3^x and h(x) = g(f(x)).

When x = 1

h(1) = f(g(1))

= f(-2)

Evaluating f(-2) using the function 3^x :

f(-2) = 3^-2

= 1/3²

= 1/9

When x = 3

h(3) = f(g(3))

= f(0)

Evaluating f(0) using the function 3^x :

f(0) = 3^-0

= 1

When x = 5

h(5) = f(g(5))

= f(1)

Evaluating f(1) using the function 3^x :

f(1) = 3¹

= 3

Question 8 :

Compare the quantity in column A with column B.

f(x) = 2x - 5 and g(x) = (1/2) (x + 5)

[A]The quantity in Column A is greater.

[B]The quantity in Column B is greater.

[C]The two quantities are equal.

[D]The relationship cannot be determined on the basis of the information supplied.

Solution :

f(x) = 2x - 5 and g(x) = (1/2) (x + 5)

Evaluating f(2) :

f(2) = 2(2) - 5

= 4 - 5

= -1

Evaluating g(f(2)) :

g(f(2)) = g(-1)

Applying x = -1 in the function g(x)

g(-1) = (1/2) (-1 + 5)

= (1/2)(4)

= 2

[B]The quantity in Column B is greater is the correct answer.

Question 9 :

Given 𝑓(𝑥)  = 5𝑥 - 2𝑏 while 𝑔(𝑥) = 4𝑏𝑥. If 𝑓[𝑔(1)] = 36, what is 𝑔(𝑓(1)) ?

Solution :

𝑓(𝑥)  = 5𝑥 - 2𝑏 while 𝑔(𝑥) = 4𝑏𝑥.

g(1) = 36

g(1) = 4b(1)

g(1) = 4b

Applying the value of g(1), we get 

f(4b) = 5 (4b) - 2b

= 20b - 2b

= 18b

36 = 18b

b = 36/18

b = 2

g(f(1)) = g(5(1) - 2b)

= g(5 - 2(2))

= g(5-4)

g(1)

Applying the value x = 1 in g(x), we get

g(1) = 4(2)(1)

= 8

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