Q. 8 Consider f : R_+\rightarrow [4,\infty) given by f (x) = x^2 + 4. Show that f is invertible with the
inverse f^{-1} of f given by f^{-1} = \sqrt{y-4} , where R_+ is the set of all non-negative
real numbers.

Answers (1)

f : R_+\rightarrow [4,\infty)

f (x) = x^2 + 4

One- one:

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

            x^{2}+4=y^{2}+4

                   x^{2}=y^{2}

\Rightarrow              x=y

  \therefore   f is one-one.

Onto:

Let for  y \in [4,\infty)  , y=x^{2}+4

                  \Rightarrow         x^{2}-y+4\geq 0

                  \Rightarrow             x^{2}\geq y-4

                \Rightarrow                  x\geq (y-4)^{\frac{1}{2}}               

\therefore for y \in [4,\infty) there is x= {\sqrt{(y-4)}}  such that 

f(x)= f(\sqrt{y-4})= \left ( \sqrt{y-4} \right )^{2}+4

                                         = y-4 +4 = y

\therefore f is onto.

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

Let  g : [4,\infty)\rightarrow R_+    by  g(y)= {\sqrt{(y-4)}}

Now,  (gof)(x)=g(f(x))=g(x^{2}+4)=\sqrt{x^{2}-4+4}=\sqrt{x^{2}}=x

        (fog)(x)=f(g(x))=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=y-4+4=y

               \therefore gof=fog=I_R

Hence,  f is invertible with the inverse f^{-1} of f given by f^{-1} = \sqrt{y-4}

 

 

 

 

 

 

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