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- 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 $f:R \to R$ defined by $\; f(x)=10x+7$

To check if a function is invertible or not ,we see if the function is both one-one and onto.

$\textbf {Step 1: Checking one-one}$

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 $f(x)=f(y) \rightarrow 10x+7=10y+y \rightarrow x = y$.

Therefore f is one-one or injective.

$\textbf {Step 2: Checking onto}$

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 $f(x) = y$.

Let $y= f(x) = 10x+7 \rightarrow x=\large \frac{y-7}{10}$

$\Rightarrow$ $f(x)=f( \large\frac{y-7}{10})=$1$0(\large \frac{y-7}{10})$$+7=y-7+7=y$

Therefore f is onto. Therefore, is invertible, since it both one-one and onto.

$\textbf {Step 3: To calculate}\; f^{-1}, \textbf {we must first define g(y):}$

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.

Let us define a function $g:R \to R$ such that $g(y)=\large \frac{y-7}{10}$

$\textbf {Step 4: Calculate gof}$

$\Rightarrow (gof)(x)=g(f(x)) =g(10x+7) =\large \frac{(10x+7)-7}{10}=\frac{10x}{10}$$=x$

$\textbf {Step 5: Calculate fog}$

$\Rightarrow (fog)(y)=f\large (\frac{y-7}{10})$$=10\large (\frac{y-7}{10})+7$ $= y$

$\textbf {Step 6: Calculating} \; f^{-} \textbf{ from} \; gof = fog$:

$gof=I_R\;and\;fog=I_R$

$\Rightarrow$ The required inverse function $f^{-1} = g:R \to R$ $g(y)=\large \frac{y-7}{10}$

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