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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

The codomain of f is the domain of g.
The domain of gof is the domain of f.
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:

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

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.

We can obtain the function

     hog : B→D

        f: A→B

Then (hog) of: A→D is defined by

     [(hog)of](x) = [hog](f(x))

                      = h[g(fx))]

So both of these functions are equal, or (hog)of = ho(gof).

Notice that in the functions (hog)of and ho(gof)

both have the same domain A
both have the same codomain D
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).

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.

(hog)of (-5) = (hog)[f(-5)]

                 = (hog)[(-5)2+2]

                 = (hog)[27]

                 = h[g(27)]

                 = h[3 *27-1]

                 = h[81-1]

                 = h(80)

                 = 1 - (80)2

                 = 1- 6400

                 = - 6399

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:

Let A = {2,4,6,8} B = {1,2,3,4} C = {11,12,13}. Let f: A→B, g: B→C 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[1]

              = 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)}

Let f: R→R and g: R→R 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)