Q. 6 Show that f : [-1, 1] \rightarrow R, given by f(x) = \frac{x}{(x + 2)} is one-one. Find the inverse of the functionf : [-1, 1] \rightarrow Range f

Answers (1)

f : [-1, 1] \rightarrow R

f(x) = \frac{x}{(x + 2)}

One -one:

            f(x)=f(y)

        \frac{x}{x+2}=\frac{y}{y+2}

     x(y+2)=y(x+2)

   xy+2x=xy+2y 

            2x=2y

             x=y

\therefore   f is one-one.

It is clear that  f : [-1, 1] \rightarrow Range f is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in  Range f\rightarrow [-1, 1]

g: Range f\rightarrow [-1, 1]

let y be an arbitrary element of range f

Since, f : [-1, 1] \rightarrow R is onto ,so

        y=f(x)  for x \in \left [ -1,1 \right ]

          y=\frac{x}{x+2}

       xy+2y=x

       2y=x-xy

     2y=x(1-y)

    x = \frac{2y}{1-y}     ,y\neq 1

 

g(y) = \frac{2y}{1-y}

 

f^{-1}=\frac{2y}{1-y},y\neq 1

 

 

 

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