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

This question appeared in 65-1.65-2 and 65-3 versions of the paper in 2013. Comment
A)
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 $f:R_+ \to [4, \infty]$ is given as $f(x) =x^2+4$
To check if a function is invertible or not, verify if the function is both one-one and onto, in which case an inverse exists..
$\textbf{Step 1: Checking one-one}$:
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 $f(x)=f(y)$ for $x,y \in R_+$.
$\Rightarrow x^2+4=y^2+4$
$\Rightarrow x^2=y^2 \rightarrow x=y$, $\;$as $x,y$ can only take up positive values.
Therefore, $f$ defined by $f(x) =x^2+4$ is a $\textbf{one-one}$ function
$\textbf{Step 2: Checking onto}$:
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 us consider an element y such that $\;y \in [4,\infty]$.
Let $y=x^2+4$
$\Rightarrow x^2=y-4 \rightarrow x=\sqrt {y-4}$
Since y is $\geq 4; \sqrt {y-4} \geq 0$
Therefore for every $y \in R$ there exists $x=\sqrt {y-4} \in R$
Therefore $f$ defined by $f(x) =x^2+4$ is an $\textbf{onto}$ function.
Therefore, since the function is both one-one and onto, it is $\textbf{invertible}$.
$\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 $g:[4,\infty] \to R\;$ by $g(y)=\sqrt {y-4}$.
To show that $g$ is inverse of $f$. we have to show $fog=I$ and $gof =I$.
$\textbf{Step 4: Calculating} \; gof$:
Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
Therefore, $(gof)(x)=g(f(x)) =g(x^2+4)$
$\Rightarrow gof =\sqrt {(x^2+4)-4} =\sqrt {x^2} =x$
$\textbf{Step 5: Calculating} \; fog$:
Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
Therefore, $(fog) (y)=f(g(y))$
$\Rightarrow fog =f(\sqrt {y-4})$ $=(\sqrt {y-4})^2+4$ $=g-4+4=y$
$\textbf{Step 6: Calculating}\; f^{-} \; \textbf{from} \;gof = fog$:
Since $gof =I_{range+}$ and $fog = I_{(4,\infty)}$, $\;f^{-1}=g$
$\Rightarrow f^{-1}(y)=y(y)=\sqrt {y-4}$
Therefore, the inverse of $f$ is $\mathbf{\sqrt {y-4}}$.