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Q. 7  Consider f : R \rightarrow R given by f (x) = 4x + 3. Show that f is invertible. Find the
inverse of f.

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f : R \rightarrow R  is given by  f (x) = 4x + 3

One-one :

Let  f(x)=f(y)

    4x + 3 = 4y+3

            4x=4y

              x=y

\therefore f is one-one function.

Onto:

y=4x+3\, \, \, , y \in R

\Rightarrow x=\frac{y-3}{4} \in R

So, for y \in R there is  x=\frac{y-3}{4} \in R   ,such that 

f(x)=f(\frac{y-3}{4})=4(\frac{y-3}{4})+3

                                    = y-3+3 

                                    = y

 \therefore f  is onto.

Thus, f is one-one and onto so f^{-1} exists.

Let, g:R\rightarrow R by g(x)=\frac{y-3}{4}

Now, 

(gof)(x)= g(f(x))= g(4x+3)

                                           =\frac{(4x+3)-3}{4}

                                          =\frac{4x}{4}

                                          =x

(fog)(x)= f(g(x))= f(\frac{y-3}{4})

                                         = 4\times \frac{y-3}{4}+3

                                         = y-3+3

                                         = y

(gof)(x)= x             and     (fog)(x)= y

Hence, function f is invertible and inverse of f is g(y)=\frac{y-3}{4}.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Posted by

seema garhwal

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