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

Hint: For $y \in R, \; f(x) = \large\frac{x}{x+2}$, for some $x \in [-1,1]$ i.e, $x = \large\frac{2y}{1-y}$

Toolbox:
• To check if a function is invertible or not ,we see if the function is both one-one and onto.
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
• Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
• A function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
Given a function $$f : R \rightarrow R$$, given by $$f(x) = 4x+3$$.
Step 1: Checking One-one.
We know that a function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
Let $f(x1)=f(x2) \Rightarrow 4x1+3 = 4x2+ 3 \rightarrow x1 = x2$.
Therefore $f$ is a one-one function.
Step 2: Checking onto:
A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
$y=f(x)$ for some $x$ $\Rightarrow y=4x+3$ $\rightarrow x =\large \frac{y-3}{4}$
$\Rightarrow f(x) = f\large (\frac{y-3}{4})$ $= 4 \large \frac{y-3}{4} + 3 = y$
Therefore, for every $y \in R$, there exisits a $x$ such that $f(x) = y$. This means that the function is on-to or surjective.
Hence the inverse of the function $f$ must exist.
Step 3: Calculating $f^{-1}$:
Since $y=f(x)$, we can define a function $g$ given by $g(y) = \large \frac{y-3}{4}$
We know that a function $g$ is called inverse of $f:x \to y$, then exists $g:y \to x$ such that $gof=I_x\;and\; fog=I_y$, where $I_x, I_y$ are identify functions.
Therefore, to check if $g$ is inverse of $f$ , we need to check if $gof=I$ and $fog=I$.
Step 4: Calculating $(gof)(x)$:
$(gof)(x)=g(f(x)) = g(4x+3) = \large \frac{4x+3-3}{4}$ $=x$
Step 5: Calculating $(fog)(x)$:
$(fog)(y)=f(g(y))=$$f\large (\frac{y-3}{4} )$ $= 4 \large \frac{y-3}{4} + 3$ $= y$
Step 6: Calculating $f^{-}$ from $gof = fog$
Since $gof =fog=I_{range+}$, $f^{-1}=g$
$\Rightarrow f^{-1}(y)=\large \frac{y-3}{4}$.
edited Mar 19, 2013