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.

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