Composition of functions -gof fog

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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}.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 R_f = D_g.

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.