Question

Let f be a real function given by $f\left(x\right)=\sqrt{x-2}$.
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f²

Also, show that fof ≠ f² .

Answer 1

$$\begin{gathered} f\left(x\right)=\sqrt{x-2} \\ \text{For domain,} \\ x-2\ge 0 \\ \Rightarrow x\ge 2 \\ \text{Domain of}f=[2,\infty ) \\ \text{Since}f\text{is a square-root function, range of}f=\left(0,\infty \right) \\ \text{So,}f:[2,\infty )\to \left(0,\infty \right) \\ \left(\mathrm{i}\right)fof \\ \text{Range of}\mathit{\text{f}}\text{is not a subset of the domain of}\mathit{\text{f}}\text{.} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\in \text{domain of}f\text{and}f\left(x\right)\in \text{domain of f}\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\in [2,\infty )\text{and}\sqrt{x-2}\in [2,\infty )\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\in [2,\infty )\text{and}\sqrt{x-2}\ge 2\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\in [2,\infty )\text{and}x-2\ge 4\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\in [2,\infty )\text{and}x\ge 6\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=}\left\{x:x\ge 6\right\} \\ \Rightarrow \text{Domain}\left(fof\right)\text{=[6, ∞)} \\ fof:[6,\; \infty )\to R \\ \left(fof\right)\left(x\right)=f\left(f\left(x\right)\right) \\ =f\left(\sqrt{x-2}\right) \\ =\sqrt{\sqrt{x-2}-2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)fofof=\left(fof\right)of \\ \text{We have,}f:[2,\infty )\to \left(0,\infty \right)\text{and}fof:[6,\infty )\to R \\ \Rightarrow \text{Range of}f\text{is not a subset of the domain of}fof. \\ \mathrm{Then},\text{domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\in \text{domain of}f\text{and}f\left(x\right)\in \text{domain of}fof\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\in [2,\infty )\text{and}\sqrt{x-2}\in [6,\infty )\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\in [2,\infty )\text{and}\sqrt{x-2}\ge 6\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\in [2,\infty )\text{and}x-2\ge 36\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\in [2,\infty )\text{and}x\ge 38\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=}\left\{x:x\ge 38\right\} \\ \Rightarrow \text{Domain}\left(\left(fof\right)of\right)\text{=[38, ∞)} \\ fof:[38,\infty )\to R \\ \mathrm{So},\left(\left(fof\right)of\right)\left(x\right)=\left(fof\right)\left(f\left(x\right)\right) \\ =\left(fof\right)\left(\sqrt{x-2}\right) \\ =\sqrt{\sqrt{\sqrt{x-2}-2}-2} \end{gathered}$$

$$\begin{gathered} \left(iii\right)\text{We have,}\left(fofof\right)\left(x\right)\text{=}\sqrt{\sqrt{\sqrt{x-2}-2}-2} \\ \mathrm{So},\left(fofof\right)\left(38\right)\text{=}\sqrt{\sqrt{\sqrt{38-2}-2}-2} \\ \text{=}\sqrt{\sqrt{\sqrt{36}-2}-2} \\ \text{=}\sqrt{\sqrt{6-2}-2} \\ \text{=}\sqrt{2-2} \\ \text{=0} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\text{We have,}fof=\sqrt{\sqrt{x-2}-2} \\ \Rightarrow f^2\left(x\right)=f\left(x\right)\times f\left(x\right)=\sqrt{x-2}\times \sqrt{x-2}=x-2 \\ \text{So,}fof\ne f^2 \end{gathered}$$