Q .9 Consider f : R_+ \rightarrow [- 5, \infty) given by f (x) = 9x^2 + 6x - 5. Show that f is invertible with f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )

Answers (1)

f : R_+ \rightarrow [- 5, \infty)

f (x) = 9x^2 + 6x - 5

One- one:

Let       f(x)=f(y)    for \, \, x,y\in R

            9x^{2}+6x-5=9y^{2}+6y-5

                   9x^{2}+6x=9y^{2}+6y

\Rightarrow              9(x^{2}-y^{2})=6(y-x)

             9(x+y)(x-y)+6(x-y)= 0

        (x-y)(9(x+y)+6)=0

Since, x and y are positive. 

         (9(x+y)+6)> 0

\therefore x=y

 \therefore   f is one-one.

Onto:

Let for  y \in [-5,\infty)  , y=9x^{2}+6x-5

                  \Rightarrow         y=(3x+1)^{2}-1-5

                  \Rightarrow             y=(3x+1)^{2}-6

                \Rightarrow                  y+6=(3x+1)^{2}

                                         (3x+1)=\sqrt{y+6}

                                           x = \frac{\sqrt{y+6}-1}{3}                                      

\therefore f is onto  and range is y \in [-5,\infty).

Since f is one-one and onto so it is invertible.

Let  g : [-5,\infty)\rightarrow R_+    by  g(y) = \frac{\sqrt{y+6}-1}{3}  

 

(gof)(x)=g(f(x))=g(9x^{2}+6x)-5=g((3x+1)^{2}-6)\\=\sqrt{{(3x+1)^{2}}-6+6} -1

       (gof)(x)=\frac{3x+1-1}{3}=\frac{3x}{3}= x        

(fog)(x)=f(g(x))=f(\frac{\sqrt{y+6}-1}{3})

                                          =[3(\frac{\sqrt{y+6}-1}{3})+1]^{2}-6

                                           =(\sqrt{y+6})^{2}-6

                                         =y+6-6

                                        =y

                                     

               \therefore gof=fog=I_R

Hence,  f is invertible with the inverse f^{-1} of f given by  f^{-1} (y) = \left (\frac{(\sqrt{y + 6}) - 1}{3} \right )

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