Q.1 Let f : R \rightarrow R be defined as f (x) = 10x + 7. Find the function g : R \rightarrow R such
that g o f = f o g = I_R.

Answers (1)

f : R \rightarrow R

f (x) = 10x + 7

g : R \rightarrow R    and    g o f = f o g = I_R

For one-one:

                   f(x)=f(y)

                10x+7=10y+7

                     10x=10y

                           x=y

   Thus, f is one-one.

For onto:

For y \in R,y=10x+7

                   x= \frac{y-7}{10} \in R

   Thus, for y \in R, there exists x= \frac{y-7}{10} \in R  such that

         f(x) = f(\frac{y-7}{10})=10(\frac{y-7}{10})+7=y-7+7=y

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let  g : R \rightarrow R   as    f(y)=\frac{y-7}{10}

gof(x)=g(f(x))= g(10x+7) =\frac{(10x+7)-7}{10}=\frac{10x}{10}=x

fog(x)=f(g(x))= f(\frac{y-7}{10}) =10\frac{y-7}{10}+7=y-7+7=y

 

\therefore             gof(x)=I_R      and   fog(x)=I_R

Hence,  g : R \rightarrow R defined as g(y)=\frac{y-7}{10}

 

 

 

    

 

 

 

 

             

 

 

 

 

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