Composite Functions

Consider three non-empty sets, as drawn below. Let f be a function from A into B and g be a function from B into C. For any a ∈ A, there exists an f (a) ∈ B.

If we take B as the domain of g, then for any f (a) ∈ B, there exists a g[f(a)] ∈ C. So g[f(a)] is defined.

We now have a rule that for every a ∈ A, the rule assigns a to a unique g[f(a)] in C. So we have a new function from A into C. This function is called the composite function of f and g and is denoted by g o f [read as “g circle f”].

Definition

Let f: A→B, g : B →C be two functions. Then the composite function of f and g denoted by gof is gof: A→C and is defined by (gof)(x) = g[ f (x)].

The composite function is also called the product function.

Notice that

1. The codomain of f is the domain of g.
2. The domain of gof is the domain of f.
3. The codomain of gof is the codomain of g.

Example 1

Let f: A→B, g : B →C be defined as given below. gof(p) = g[f(p)] = g(a) = 6

gof(q) = g[f(q)] = g(b) = 2

gof(r) = g[f(r)] = g(b) = 2

Example 2

Let f = {(1,2), (2,3), (3,4)} g = {(2,1), (3,2), (4,4)}. Find gof.

gof(1) = g[f(1)] = g(2) = 1

gof(2) = g[f(2)] = g(3) = 2

gof(3) = g[f(3)] = g(4) = 4 gof = { (1,1), (2,2), (3,4)}

Can we find fog ?

(fog)(x) = f(g(x))

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

f(g(3)) = f(2) = 1

f(g(4)) = f(4) which is undefined.

So fog is not defined.

Example 3

Let f: R→R, g:R→R defined by f(x) = 3x-2 g(x) = x - 2/3. Find gof and fog.

Both gof and fog are defined.

(fog)(x) = f[g(x)] = f(x-2/3]

= 3(x - 2/3) - 2

= 3x - 3*2/3 - 2

= 3x-4

(gof)(x) = g[f(x)] = g[3x-2]

= 3x-2-2/3

= 3x - (6+2/3)

= 3x - 8/3

We can see that

fog ≠ gof.

Example 4

Let f(x) = x-2 g(x) = x2+1 where f:R→R and g:R→R.

Find    i) gof(-4), ii) (fog)(-4), iii) Give rules for fog and gof.

Solutions:

i)    gof (-4)

= g[f(-4)]

= g[-4-2]

= g(-6)

= (-6)2+1

= 36+1

= 37

ii)    fog(-4) = f[g(-4)]

= f[(-4)2 + 1]

= f[16+1]

= f(17)

= 17-2

= 15

iii)    fog(x) = f[g(x)] = f(x2+1)

= (x2+1) - 2

= x2+1-2

= x2-1

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

= (x-2)2 + 1

= x2 - 4x + 2 + 1

= x2 - 4x + 3

= x2 - 4x + 2 + 1

= x2 - 4x + 3

Consider the following figure. Let f: A→B, g: B→C, h: C→D

We can obtain the composite function

gof: A→C

h : C→D

We can combine them to get the function h[gof(x)] = (hogof)(x) or hogof(x) = h[g(f(x))]

Now consider this figure.

1. both have the same domain A
2. both have the same codomain D
3. for every x ∈A, there exists the same image h[g(f(x))] in D.

Example 5

Let f: R→R, g: R→R, h: R→R. If f(x) = x2+2, g(x) = 3x-1, h(x) = 1-x2.

Find a) ho(gof)(-5) b) (hog)of (-5) c) rules for ho(gof).

1. ho(gof)(-5) = ho[g(f (-5)]
= ho[g((-5)2+2)]
= ho[g(25+2)]
= ho[g(27)]
= ho[3*27-1]
= ho[81-1]
= h ((80))
= 1 - (80)2
= 1 - 6400
= - 6399.

2. (hog)of (-5) = (hog)[f(-5)]
= (hog)[(-5)2+2]
= (hog)
= h[g(27)]
= h[3 *27-1]
= h[81-1]
= h(80)
= 1 - (80)2
= 1- 6400
= - 6399

3. ho(gof) (x) = ho[g(f(x))]
= ho[g(x2+2)]
= h[3(x2+2)-1]
= h[3x2+6-1]
= h[3x2+5]
= 1-(3x2+5)2
= 1-(9x4+30x2+25)
= 1-9x4-30x2-25
= -9x4-30x2-24

Try these questions:

1. Let  A = {2,4,6,8}  B = {1,2,3,4}  C = {11,12,13}. Let f: AB, g: BC be defined as
f = {(2,2), (4,1), (6,4), (8,3)}       g = {(1,11), {2,12), (3,13), (4,11)}
Find gof.
Solution:
A = {2,4,6,8}    B = {1,2,3,4}    C = {11,12,13}.
f = {(2,2), (4,1), (6,4), (8,3)}
g = {(1,11), {2,12), (3,13), (4,11)}
To find gof
gof(2) = g[f(2)]
= g(2) = 12
gof(4) = g[f(4)]
= g
= 11
gof(6) = g[f(6)]
= g(4)
= 11
gof(8) = g[f(8)]
= g(3)
= 13
gof = { (2,12), (4,11), (6,11), (8,13)}

2. Let f: RR and g: RR be defined by f(x) = 1-5x / 2 g(x) = 3x+4.  Find
a.   fog(4)
b.   gof(4)
c.   fog(x)
d.   gof(x)
Solutions:
f: R →R g : R →R defined by f(x) = 1- 5x / 2, g (x) = x/3x + 4

a.   fog(4) = f[g(4)] b.   gof(4) = g[f(4)] c.   fog(x) = f[g(x)] d.   gof(x) = g[f(x)] 