# Consider $$f:R_+ \to [\;\text{–5}, \infty )$$given by $$f(x)=9x^2 +6x$$-$$5$$.Show that $$f$$ is invertible with $$f^{-1} (y) = \bigg(\frac{(\sqrt{y+6}) -1} { 3}\bigg)$$

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 function $f:R_+ \to [-5, \infty)$ given by $f(x) =9x^2+6x-5$
To check if a function is invertible or not, we see if the function is both one-one and onto.
$\textbf {Step 1: Calculating one-one}$
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 $x,y \in R_+$.
$\Rightarrow 9x^2+6x-5=9y^2+6y-5 \rightarrow$ $9x^2+6x=9y^2+6y \rightarrow x=y$
Therefore $f$ is a one-one function
$\textbf {Step 2: Calculating onto}$
Let $y$ be an element in $[-5,\infty)$ such that $y=9x^2+6x-5$
$\Rightarrow y=(3x+1)^2-1-5=(3x+1)^2-6$
$\Rightarrow 3x+1=\sqrt {y+6} \rightarrow x=\large \frac{(\sqrt {y+6})-1}{3}$$; x \in R_+$
As $y \geq {-5} => y+6 \geq 0$, $f$ is onto.
Hence the inverse of the function $f$ must exist.
$\textbf {Step 3: To calculate the inverse of } f, \textbf {we must first define} \;g(y):$
Let us define a function $g:[-5,\infty) \to R_+$ given by $g(y)=\large \frac{\sqrt {y+6}-1}{3}$
$\textbf {Step 4: Calculating}\; gof$:
$(gof)(x)=g(f(x))=g(9x^2+6x-5) = 9x^2+6x-5=(3x+1)^2-6$
$\Rightarrow$ $(gof)(x)=\Large \frac{\sqrt {(3x+1)^2-6+6}-1}{3}$
$\Rightarrow (gof)$ $=\Large \frac {3x+1-1}{3}$ $=x$
$\textbf {Step 5: Calculating}\; fog$:
$(fog) (y)=f(g(y))$ $=f(\large \frac{\sqrt {y+6}-1}{3})$ $=[3(\large \frac{\sqrt {y+6}-1}{3})+1]^2$
$\Rightarrow fog$ $=[(\sqrt {y+6}-1+1]^2-6$ $=(\sqrt {y+6})^2-6 = y$
$\textbf {Step 6: Calculating} \;\;f^{-}\; \textbf {from}\; \;gof = fog$
From Steps 4 and 5, we can see that $gof=I_R\; = \;fog=I_{[-5,\infty)}$
Hence $f_{-1}$ is given by $f^{-1}(y)=g(y)=\Large \frac{\sqrt {y+6}-1}{3}$
edited Mar 19, 2013