# Constructions of functions

• Nov 9th 2011, 03:29 PM
Constructions of functions
So here's the question:
Find f•g (relation) for each pair of functions f and g. Use understood domains for f and g.

a.) f(x)=2x+5, g(x)=6-7x

(f•g)(x)= 2(6-7x)+5 = 17-14x

(g•f)(x)= 6+7(2x+5) = -14x-29

Okay i understand this. But

b.) f(x)={(t,r),(s,r),(k,l)} , g(x)={(k,s),(t,s),(s,k)}

f•g = {(k,r),(t,r),(s,l)} but g•f=ø (null set)

i see how f•g works, replace the x value of f with the x value of g. But why is the relation of g•f the null set?
wouldnt it be {(t,s),(s,s),(k,k)}?
Now i see that t and s have the same y variable 's' which doesent make this a function.
BUT f(x) has t and s with the same value 'r' so its not a function in the first place right??

• Nov 9th 2011, 03:59 PM
Sudharaka
Re: Constructions of functions
Quote:

So here's the question:
Find f•g (relation) for each pair of functions f and g. Use understood domains for f and g.

a.) f(x)=2x+5, g(x)=6-7x

(f•g)(x)= 2(6-7x)+5 = 17-14x

(g•f)(x)= 6+7(2x+5) = -14x-29

Okay i understand this. But

b.) f(x)={(t,r),(s,r),(k,l)} , g(x)={(k,s),(t,s),(s,k)}

f•g = {(k,r),(t,r),(s,l)} but g•f=ø (null set)

i see how f•g works, replace the x value of f with the x value of g. But why is the relation of g•f the null set?
wouldnt it be {(t,s),(s,s),(k,k)}?
Now i see that t and s have the same y variable 's' which doesent make this a function.
BUT f(x) has t and s with the same value 'r' so its not a function in the first place right??

$\displaystyle g\circ f (x)= g(f (x))$

So domain of the function $\displaystyle g\circ f$ is, $\displaystyle \left\{t, s, k\right\}$.

If you take the element t, it maps to r under f, so;

$\displaystyle g\circ f (t)= g(f (t)) = g(r)$

But the element r is not included in the domain of g; hence, g(r) is not defined. Similar situation occurs for both s and k as well. Hence the set of ordered pairs of $\displaystyle g\circ f$ is empty.

i.e: $\displaystyle g\circ f (x)= g(f (x)) = \emptyset$
• Nov 9th 2011, 06:29 PM
Re: Constructions of functions
I'm still confused. By using that idea wouldn't f•g be empty as well? there is no 'l' or 'r' in g(x) either.
And i don't know what you mean by "maps."
• Nov 9th 2011, 07:14 PM
Deveno
Re: Constructions of functions
let's look at f and g each as a series of "links". so when we say f(k) = l, we mean f takes k as a source, and links it to a target, l.

here's f:

k-->l
s-->r
t-->r the fact that t and k both go to r, does not violate the definition of a function, for example the squaring function takes both -2 and 2 to 4. if k went to r and l, THAT would be a problem.

now, here's g:

k-->s
t-->s
s-->k

now fog (or "f follows g") means we start with possible starting places for g. if any of those end up at a starting place for f, we can continue:

..g.....f....

k-->s-->r (g takes k to s, f takes s to r).
t-->s-->r (g takes t to s, f takes s to r).
s-->k-->l (g takes s to k, f takes k to l).

this means fog = {(k,r), (t,r), (s,l)}

what happens when we try to define gof ("g follows f")?

...f....g....

k-->l-->?
s-->r-->?
t-->r-->? neither l nor r is a "starting point" for g, so we don't have any target defined for gof(k), gof(s) or gof(t).
• Nov 9th 2011, 08:39 PM