Twelve Standard >> Composition of functions | Illustration-2

Composition of functions : Illustration

Example:
Let f:N \(\rightarrow\) N, g:N \(\rightarrow\) N and h:N \(\rightarrow\) N be the functions defined by f(x)=3x, f(y)=+4 and h(z)=\(z^{2}\), for all x, y, z \(\in\) N. Show that ho(gof)=(hog)of

Solve:
f, g, h: N \(\rightarrow\) N, gof, hog, ho(gof) & (hog)of can be defined.
gof=gof(x)=g(f(x))=f(x)+4
     =3x+4
ho(gof)=\((gof)^{2}\)
           =\((3x+4)^{2}\)....(1)
hog=h(g(x))
      =\((g(x))^{2}\)=\((y+4)^{2}\)
(hog)of=\((f(x)+4)^{2}\)
           =\((3x+4)^{2}\)....(2)
\(\therefore\) ho(gof)= (hog)of   Proved

Example:
If f:R \(\rightarrow\) R s a function defined by f(x)=ax+b for all x \(\in\) R, then find the value of a and b so that fof=\(I_R\)

Solve:
    fof=\(I_R\)
\(\Rightarrow\) fof(x)=x
\(\Rightarrow\) f(f(x))=x
\(\Rightarrow\) af(x)+b=x
\(\Rightarrow\) a(ax+b)+b=x
\(\Rightarrow\) \(a^{2}\)x+ (a+1)b=x+0
Because coefficient of x in righthand side is 1
\(\Rightarrow\) \(a^{2}\)=1,
\(\Rightarrow\) a=\(\pm1\)
Constant term in righthand side is 0
\(\therefore\) (a+1)b=0
if put a=1 in this equation then,
   b=0
if put a=-1 in this equation then,
  (-1+1)b=0
\(\Rightarrow\) 0.b=0
\(\Rightarrow\) b be any real number

So solution is a=1,b=0
    and a=-1, b\(\in\) R