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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 function $f : [-1, 1] \rightarrow Range f$

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$f : [-1, 1] \rightarrow R$

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

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 $x \in \left [ -1,1 \right ]$

$y=\frac{x}{x+2}$

$xy+2y=x$

$2y=x-xy$

$2y=x(1-y)$

$x = \frac{2y}{1-y}$ , $y\neq 1$

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

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

Posted by

seema garhwal

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