Discrete Math For each of the following pairs of functions f

Discrete Math:

For each of the following pairs of functions f and g, compute g o f and f o g if the compositions exist. If either of the compositions does not exist, then say so. f(x) = 2x + 1 and g(x) = x + 3 for all x epsilon Z. f = {(1, 2), (3, 4), (5, 6)} and g = {(2, 1), (4, 3), (6, 5)} f = id_A and g = id_B, where A B

Solution

Not that for (fog) to exist range of g must be subset of domain of f

This is easy to see why.

(fog)(x)=f(g(x)). So we see f acts on range of g. And it must be defined at g(x). HEnce range of g must be subset of f

1.

(fog)(x)=f(g(x))=2(x+3)+1=2x+7

(gof)(x)=g(f(x))=2x+1+3=2x+4

2.

First compute: (fog)(x)

g(x)={1,3,5} ie range of g

This is exactly the domain set of f

So to write down (fog) in set form as in problem.

2 maps to 1 under g. Then 1 maps to 2 under f

4 maps to 3 under g. Then 3 maps to 3 under f

6 maps to 5 under g. Then 5 maps to 6 under f

fog={(2,2),(4,3),(6,6)}

First compute: (gof)(x)

f(x)={2,4,6} ie range of f

This is exactly the domain set of g

So to write down (gof) in set form as in problem.

1 maps to 2 under f. Then 2 maps to 1 under g

3 maps to 4 under f. Then 4 maps to 3 under g

5 maps to 6 under f. Then 6 maps to 5 under g

gof={(1,1),(3,3),(5,5)}

3.

First we compute: fog

Range of g is B . But domain of f is A. And it is given A is not equal to B but it is a proper subset of B.

Hence, fog is not defined.

Now we compute:gof

This is defined as : (range of f=A) is a subset of domain of g

(gof)(x)=g(f(x))=g(x)=x=id_A