Consider f : R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with

Injectivity of $f$ :

Let $x$ and $y$ be two elements of domain $\left(R^{+}\right)$, such that

$f(x)=f(y)$

$\Rightarrow 9 x^{2}+6 x-5=9 y^{2}+6 y-5$

$\Rightarrow 9 x^{2}+6 x=9 y^{2}+6 y$

$\Rightarrow x=y\left(\right.$ As $\left., x, y \in R^{+}\right)$

So, $f$ is one-one.

Surjectivity of f:
Let y is in the co domain (Q) such that f(x) = y

$\Rightarrow 9 x^{2}+6 x-5=y$

$\Rightarrow 9 x^{2}+6 x=y+5$

$\Rightarrow 9 x^{2}+6 x+1=y+6$ (Adding 1 on both sides)

$\Rightarrow(3 x+1)^{2}=y+6$

$\Rightarrow 3 x+1=\sqrt{y+6}$

$\Rightarrow 3 x=\sqrt{y+6}-1$

$\Rightarrow x=\frac{\sqrt{y+6}-1}{3} \in R^{+}$(domain)

$\Rightarrow f$ is onto.

So, $f$ is a bijection and hence, it is invertible.

Finding $f^{-1}$ :

Let $f^{-1}(x)=y$               ....(1)

$\Rightarrow x=f(y)$

$\Rightarrow x=9 y^{2}+6 y-5$

$\Rightarrow x+5=9 y^{2}+6 y$

$\Rightarrow x+6=9 y^{2}+6 y+1 \quad$ (adding 1 on both sides)

$\Rightarrow x+6=(3 y+1)^{2}$

$\Rightarrow 3 y+1=\sqrt{x+6}$

$\Rightarrow 3 y=\sqrt{x+6}-1$

$\Rightarrow y=\frac{\sqrt{x+6-1}}{3}$

So, $f^{-1}(x)=\frac{\sqrt{x+6-1}}{3} \quad[$ from (1) $]$