COMPOSITION OF FUNCTIONS EXAMPLES

Example 1 :

If f(x) = 2x² + 3 and g(x) = x + 2, find f o g.

Solution :

f o g = f[g(x)]

= f[x + 2]

= 2(x + 2)² + 3

= 2(x² + 4x + 4) + 3

= 2x² + 8x + 8 + 3

= 2x² + 8x + 11

Example 2 :

If f(x) = 5x and g(x) = x + 2, find f o g(x²).

Solution :

f o g(x²) = f[g(x²)]

= f[x² + 2]

= 5(x² + 2)

= 5x² + 10

Example 3 :

If f(x) = 5x + 3 and g(x) = 7x - 2, find f o g(3).

Solution :

 f o g(x) = f[g(x)]

= f[7x - 2]

= 5(7x - 2) + 3

= 35x - 10 + 3

f o g(x) = 35x - 7

f o g(3) = 35(3) - 7

= 105 - 7

= 98

Example 4 :

Using f(x) = 4x + 3 and g(x) = x - 2, find f[g(5)].

Solution :

f[g(x)] = f[x - 2]

= 4(x - 2) + 3

= 4x - 8 + 3

    f[g(x)] = 4x - 5

f[g(5) = 4(5) - 5

= 20 - 5

= 15

Example 5 :

Using f(x) = 6x² and g(x) = 14x + 4, find g o f.

Solution :

g o f = g[f(x)]

= g[6x²]

= 84x² + 4

Example 6 :

Using f(x) = 5x + 4 and g(x) = x - 3, find g o f(6).

Solution :

g o f(x) = g[f(x)]

= g[5x + 4]

= (5x + 4) - 3

= 5x + 4 - 3

g o f(x) = 5x + 1

g o f(6) = 5(6) + 1

= 30 + 1

= 31

Example 7 :

If f(x) = x - 5 and g(x) = 2x + 3, verify f o g = g o f.

Solution :

f o g = [g(x)]

= f[2x + 3]

= (2x + 3) - 5

= 2x + 3 - 5

f o g = 2x - 2 ---->(1)

g o f = g[f(x)]

= g[x - 5]

= 2(x - 5) + 3

= 2x - 10 + 3

= 2x - 7 ---->(2)

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

Example 8 :

Let f(x) = x + k and g(x) = 7x. If  f o g(2) = 7, find k. 

Solution :

f o g(2) = 7

f[g(2)] = 7

f[7(2)] = 7

f(14) = 7

14 + k = 7

k = -7

Example 9 :

Use 𝑓 and 𝑔 by the following table of values to evaluate the following

Solution :

a) In the first table, when x = 0, the output is 1. So,

𝑓(0) = 1

b) In the second table, when input = 1 the output is 2. So,

𝑔(1) = 2

c) (𝑓◦𝑔)(−1)

(𝑓◦𝑔)(−1) = f[g(-1)] ----(1)

Evaluating g(-1) :

g(-1) = 3

Applying the value of g(-1) in (1), we get

(𝑓◦𝑔)(−1) = f(3)

(𝑓◦𝑔)(−1) = 4

So, the value of (𝑓◦𝑔)(−1) is 4.

d) (𝑓◦𝑔)(4)

(𝑓◦𝑔)(4) = f(g(4)) ----(1)

Evaluating g(4) :

From second table, the value of g(4) is -2.

Applying the value of g(4) in (1), we get

(𝑓◦𝑔)(4) = f(-2)

Evaluating f(-2) :

f(-2) = 0

So, the value of (𝑓◦𝑔)(4) is 0.

e) (𝑔◦𝑓)(0)

(𝑔◦𝑓)(0) = g[f(0)]

Applying the value of f(0),

= g[1]

Applying the value of g(1), we get

= 2

So, the value of (𝑔◦𝑓)(0) is 2.

f) (𝑔◦𝑓)(7)

(𝑔◦𝑓)(7) = g[f(7)]

Applying the value of f(7)

= g(6)

The value of g(6) from second table, g(6) = -4

So, the value of (𝑔◦𝑓)(7) is -4.

g) (𝑓◦𝑔)(1)

(𝑓◦𝑔)(1) = f[g(1)]

Applying the value g(1) from the table.

= f[2]

The value of f(2) does not exists.

h) (𝑔◦𝑓)(−2)

(𝑔◦𝑓)(−2) = 𝑔[𝑓(−2)]

Applying the value f(-2) from table, we get f(-2) = 0

= g(0)

Finding the value of g(0), we get no result for 0 as input. Then the value of g(0) does not exists.

Example 10 :

If 𝑓 = {(3, 4),(4, 5),(5, 6),(6, 7)} and 𝑔 = {(5, 3),(6, 4),(7, −2), (8, 0)}, determine:

a) (𝑓◦𝑔)(𝑥)

b) (𝑔◦𝑓)(𝑥)

Solution :

𝑓 = {(3, 4),(4, 5),(5, 6),(6, 7)} and 𝑔 = {(5, 3),(6, 4),(7, −2), (8, 0)}

a) (𝑓◦𝑔)(𝑥)

For the function g, the inputs are 5, 6, 7 and 8

When x = 5

(𝑓◦𝑔)(𝑥) = (𝑓◦𝑔)(5)

= f[g(5)] -----(1)

From the relation g :

g(5) = 3

= f(3)

The value of f(3) is 4. Then the answer for (𝑓◦𝑔)(5) is 4.

When x = 6

(𝑓◦𝑔)(𝑥) = (𝑓◦𝑔)(6)

= f[g(6)] -----(1)

From the relation g :

g(6) = 4

= f(4)

The value of f(4) is 5. Then the answer for (𝑓◦𝑔)(6) is 5.

When x = 7

(𝑓◦𝑔)(𝑥) = (𝑓◦𝑔)(7)

= f[g(7)] -----(1)

From the relation g :

g(7) = -2

= f(-2)

The value of f(-2) is none. Then the answer for (𝑓◦𝑔)(7) does not exists.

When x = 8

(𝑓◦𝑔)(𝑥) = (𝑓◦𝑔)(8)

= f[g(8)] -----(1)

From the relation g :

g(8) = 0

= f(0)

The value of f(0) is none. Then the answer for (𝑓◦𝑔)(8) does not exists.

b) (𝑔◦𝑓)(𝑥)

(𝑔◦𝑓)(𝑥) = g[f(x)]

𝑓 = {(3, 4),(4, 5),(5, 6),(6, 7)} and 𝑔 = {(5, 3),(6, 4),(7, −2), (8, 0)}

For the function f, the inputs are 3, 4, 5 and 6.

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