Question

Consider f : R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with $f^{-1}\left(x\right)=\frac{\sqrt{x+6}-1}{3}$.

Answer 1

Injectivity of f :
Let x and y be two elements of domain (R⁺), such that
f(x)=f(y)
$$\begin{gathered} \Rightarrow 9x^2+6x-5=9y^2+6y-5 \\ \Rightarrow 9x^2+6x=9y^2+6y \\ \Rightarrow x=y\left(\text{As,}x,y\in R^{+}\right) \end{gathered}$$
So, f is one-one.

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

$$\begin{gathered} \Rightarrow 9x^2+6x-5=y \\ \Rightarrow 9x^2+6x=y+5 \\ \Rightarrow 9x^2+6x+1=y+6\left(\text{Adding 1 on both sides}\right) \\ \Rightarrow {\left(3x+1\right)}^2=y+6 \\ \Rightarrow 3x+1=\sqrt{y+6} \\ \Rightarrow 3x=\sqrt{y+6}-1 \\ \Rightarrow x=\frac{\sqrt{y+6}-1}{3}\in R^{+}\left(\text{domain}\right) \end{gathered}$$

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

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=9y^2+6y-5 \\ \Rightarrow x+5=9y^2+6y \\ \Rightarrow x+6=9y^2+6y+1\left(\text{adding 1 on both sides}\right) \\ \Rightarrow x+6={\left(3y+1\right)}^2 \\ \Rightarrow 3y+1=\sqrt{x+6} \\ \Rightarrow 3y=\sqrt{x+6}-1 \\ \Rightarrow y=\frac{\sqrt{x+6}-1}{3} \\ \text{So,}{f}^{-1}\left(x\right)=\frac{\sqrt{x+6}-1}{3}[\text{from}\left(1\right)] \end{gathered}$$