Question

Let f : [−1, ∞) → [−1, ∞) be given by f(x) = (x + 1)² − 1, x ≥ −1. Show that f is invertible. Also, find the set S = {x : f(x) = f⁻¹ (x)}.

Answer 1

$$\begin{gathered} \mathrm{Injectivity}:\mathrm{Let}\mathrm{x}\mathrm{and}\mathrm{y}\in [-1,\infty ),\mathrm{such}\mathrm{that} \\ f\left(x\right)=f\left(y\right) \\ \Rightarrow {\left(x+1\right)}^2-1={\left(y+1\right)}^2-1 \\ \Rightarrow {\left(x+1\right)}^2={\left(y+1\right)}^2 \\ \Rightarrow \left(x+1\right)=\left(y+1\right) \\ \Rightarrow x=y \\ \mathrm{So},f\mathrm{is}\mathrm{a}\mathrm{injection}. \\ \mathrm{Surjectivity}:\mathrm{Let}\mathrm{y}\in [-1,\infty ). \\ \mathrm{Then},\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y} \\ \Rightarrow {\left(\mathrm{x}+1\right)}^2-1=\mathrm{y} \\ \Rightarrow \mathrm{x}+1=\sqrt{\mathrm{y}+1} \\ \Rightarrow \mathrm{x}=\sqrt{\mathrm{y}+1}-1 \\ \mathrm{Clearly},\mathrm{x}=\sqrt{\mathrm{y}+1}-1\mathrm{is}\mathrm{real}\mathrm{for}\mathrm{all}\mathrm{y}\ge -1. \\ \mathrm{Thus},\mathrm{every}\mathrm{element}\mathrm{y}\in [-1,\infty )\mathrm{has}\mathrm{its}\mathrm{pre}-\mathrm{image}\mathrm{x}\in [-1,\infty )\mathrm{given}\mathrm{by}\mathrm{x}=\sqrt{\mathrm{y}+1}-1. \\ \Rightarrow f\mathrm{is}\mathrm{a}\mathrm{surjection}. \\ \mathrm{So},f\mathrm{is}\mathrm{a}\mathrm{bijection}. \\ \mathrm{Hence},f\mathrm{is}\mathrm{invertible}. \\ \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow f\left(y\right)=x \\ \Rightarrow {\left(y+1\right)}^2-1=x \\ \Rightarrow {\left(y+1\right)}^2=x+1 \\ \Rightarrow y+1=\sqrt{x+1} \\ \Rightarrow y=\pm \sqrt{x+1}-1 \\ \Rightarrow f^{-1}\left(x\right)=\pm \sqrt{x+1}-1[\text{from}\left(1\right)\text{]} \\ f\left(x\right)=f^{-1}\left(x\right) \\ \Rightarrow {\left(x+1\right)}^2-1=\pm \sqrt{x+1}-1 \\ \Rightarrow {\left(x+1\right)}^2=\pm \sqrt{x+1} \\ \Rightarrow {\left(x+1\right)}^4=x+1 \\ \Rightarrow \left(x+1\right)\left[{\left(x+1\right)}^3-1\right]=0 \\ \Rightarrow x+1=0\text{or}{\left(x+1\right)}^3-=0 \\ \Rightarrow x=-1\text{or}{\left(x+1\right)}^3\text{=1} \\ \Rightarrow x=-1\text{or}\mathit{\text{x}}\text{+1=1} \\ \Rightarrow x=-1\text{or x=0} \\ \Rightarrow S=\left\{0,-1\right\} \end{gathered}$$