Relations and Functions


   
 
Composition of two functions or Product of two functions
Let f:AgB and g:AgB be two functions. Thus the composition of two functions f and g denoted by gof or fog is the function from A into C defined by gof = {(a,b) for some c Î B, (a,c) Î f and (c,b) Î g}.
 
 
Remark:
 
gof is defined only if Rf = Dg.
 
 
Example 1:
 
 
Suggested answer:
 
 
 
 
Example 2:
 
If f : R→R and g: R→R denoted by f(x) = x + 2, and g(x) = 2x2 + 7. Compute fog and gof.
 
Suggested answer:
 
 
 
 
Theorem
 
Statement:
 
f : S T, g : T U and h : U V be function then
 
a) (hog)of = ho(gof).
 
b) If each f and g is one-to-one then so is gof.
 
c) If each f and g is onto then so is gof.
 
 
Proof:
 
a) (hog)of = ho(gof)
 
Since, (hog)of and ho(gof) are functions from S to V. We have to prove that for any the image of S under ho(gof) = image of x under (hog)of.
 
 
 
 
 
b) If each f and g is one-to-one, then so is gof.
 
 
 
 
 
 
Hence gof is one-one .
 
c) If each f and g is also then so is gof.
 
 

To prove gof is onto we have to prove that every element of u Î U is an image element for some x Î S under gof, since g is onto t Î T such that g(t) = u.

 
 
gof(x) = g(f(x)) = g(t) = u
 
 
 
 
gof : S g U is a function .
 
gof : S g U is one-one and onto by (b) and (c).
 
(gof)-1 exists and (gof)-1 : U®S is also a one-one and onto function.
 
Again f-1and y-1 exists and one-one and onto because f and g are one-one and onto by (b) and (c).
 
 
 
 
( if is one-one and onto)
 
 
 
Further, we have
 
 
= f-1 (t) =  x       ...(ii)
 
From (i) and (ii), we get
 
 
Corollary:
 
The function f : A g B is one-one and onto.
 
 
      
 
Proof:
 
f : A g B and is one-one and onto.
 
\ for x Î A , we have an element b Î B such that
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
This proves that f is one-one and onto.
 
 
     
   
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