FINDING COMPOSITION OF FUNCTIONS

Example 1 :

If f(x) = 3x - 5 and g(x) = x², find (f∘g) (x).

Solution :

(f∘g) (x) = f ( g(x) )

Here the value of g(x) is x².

(f∘g) (x) = f ( x² )

f (x²) looks like f(x), here x = x²

So, in the function f(x) put x², where we see x.

(f∘g) (x) = 3x² - 5

Example 2 :

Let f(x) = -3x + 7 and g(x) = 2x² - 8, find (f∘g) (x) and (g∘f) (x).

Solution :

(i)  f(x) = -3x + 7 and g(x) = 2x² - 8

(f∘g) (x) = f (g(x) )

= f(2x² - 8)

Now in f(x), x can be replaced by 2x² - 8.

f(2x² - 8) = -3(2x² - 8) + 7

Distributing -3 and combining like terms, we get

= -6x² + 24 + 7

= -6x² + 31

(ii)  (g∘f) (x) = g( f(x) )

= g(-3x + 7)

Now in g(x), x can be replaced by -3x + 7

g(-3x + 7) = 2(-3x + 7)² - 8

Using the algebraic identity, expanding it.

g(-3x + 7) = 2(9x² - 42x + 49) - 8

= 18x² - 84x + 98 - 8

= 18x² - 84x + 90

Example 3 :

If f(x) = -9x - 9 and g(x) = √(x - 9), find (f∘g) (x).

Solution :

(f∘g) (x) = f( g(x) )

= f( g(x) )

Example 4 :

If f(x) = -2x + 1 and g(x) = √(x² - 9), find

(i)  (f∘g) (x) and 

(ii)  (g∘f) (x)

Solution :

Finding (f∘g) (x) :

Finding (g∘f) (x) :

Example 5 :

If f(x) = -2x + 1 and g(x) = √(x² - 5), find

(i)  (f∘g) (x) and 

(ii)  (g∘f) (x)

Solution :

(i)  Finding (f∘g) (x) :

(f∘g) (x) = f ( g(x) )

Here the value of g(x) is √(x² - 5).

= f(√(x² - 5))

In the function f(x), we will apply x as √(x² - 5).

f(g(x)) = -2(x² - 5) + 1

= -2x² + 10 + 1

= -2x² + 11

(ii)  Finding (g∘f) (x) :

(g∘f) (x) = g ( f(x) )

Here the value of f(x) is -2x + 1.

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

Here g(-2x + 1) exactly matches with g(x). So, apply x as -2x + 1 in the function g(x).

=  √((-2x + 1)² - 5)

=  √(4x² - 4x + 1 - 5)

=  √(4x² - 4x - 4)

=  √4(x² - x - 1)

=  2√(x² - x - 1)