# EVALUATING COMPOSTIE FUNCTIONS

The composition of functions is the process of combining two or more functions into a single function. The symbol of the composition of functions is ∘.

Let us see an example to see how to find composition of two functions.

Let the two functions be

f(x) = 2x - 7 and g(x) = 3x + 5

find (f∘g)(x).

Finding (f∘g)(x) :

(f∘g)(x) = f[g(x)]

= f[3x + 5]

We consider this function as f(x). Instead of x we have 3x + 5. So, in the function f(x), replace x by 3x + 5.

f(3x + 5) = 2(3x + 5) - 7

= 6x + 10 - 7

= 6x + 3

Example 1 :

Given f(t) = t2 - t and h(x) = 3x + 2 , evaluate (f ∘ h)(1).

Solution :

(f ∘ h)(1) = f[h(1)] ----(1)

Let us evaluate h(1).

h(x) = 3x + 2

h(1) = 3(1) + 2

h(1) = 5

By applying the value of h(1) = 5 in (1) , we get

(f ∘ h)(1) = f(5)

Finding the value of f(5), we get

f(5) = 52 - 5

f(5) = 25 - 5

f(5) = 20

Example 2 :

Given each pair of functions

f(x) = 4x + 8, g(x) = 7 - x2

calculate

(i)  f (g (0)) and (ii)  g ( f (0))

Solution :

f(x) = 4x + 8, g(x) = 7 - x2

Evaluating  f (g (0)) :

(i)  f (g (0))  -----(1)

g(0) = 7 - 02

g(0) = 7

Applying g(0) = 7 in (1), we get

f(g(0)) = f(7)

f(7) = 4(7) + 8

f(7) = 28 + 8

f(7) = 36

(ii) Evaluating g (f (0)) :

g ( f (0)) ----(2)

f(0) = 4(0) + 8

f(0) = 8

Applying f(0) = 8 in (2), we get

g(f(0)) = g(8)

g(8) = 7 - 82

= 7 - 64

= -57

Example 3 :

Given each pair of functions

f(x) = 5x + 7, g(x) = 4 - 2x2

calculate

(i)  f (g (0)) and (ii)  g (f (0))

Solution :

Evaluating  f (g (0)) :

(i)  f (g (0))  -----(1)

g(0) = 4 - 2(0)2

g(0) = 4

Applying g(0) = 4 in (1), we get

f(g(0)) = f(4)

f(4) = 5(4) + 7

f(4) = 20 + 7

f(4) = 27

(ii) Evaluating g (f (0)) :

g ( f (0)) ----(2)

f(0) = 5(0) + 7

f(0) = 7

Applying f(0) = 7 in (2), we get

g(f(0)) = g(7)

g(7) = 4 - 2(72)

= 4 - 2(49)

= 4 - 98

= -94

Example 4 :

Given that f(x) = 3x - 4 and g(x) = 7 - xcalculate.

(i) (f ∘ g)(2)

(ii)  (g ∘ f)(2)

(iii)  (f ∘ g)(-2)

(iv)  (g ∘ f)(-2)

Solution :

(i) (f ∘ g)(2) = f[g(2)]

 g(2) = 7 - 22= 7 - 4= 3 f[g(2)] = f(3)= 3(3) - 4= 9 - 4(f ∘ g)(2) = 5

(ii)  (g ∘ f)(2) = g[f(2)]

= g[3(2) - 4]

= g[6 - 4]

= g[2]

Now in the function g(x), instead of x we will apply 2.

7 - 2

= 7 - 4

(g ∘ f)(2) = 3

(iii)  (f ∘ g)(-2) = f[g(-2)]

g(x) = 7 - x2

= f [7 - (-2)2]

= f [7 - 4]

= f [3]

Now in the function f(x), we will replace x by 3.

f(x) = 3x - 4

= 3(3) - 4

= 9 - 4

= 5

(iv)  (g ∘ g)(-2) = g[g(-2)]

g[g(2)] = g[7 - (-2)2

= g[7 - 4]

= g[3]

Now in the function g(x), instead of x we will apply 3.

g(3) = 7 - 32

= 7 - 9

= -2

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