Let $f$ be a real function given by $f(x)=\sqrt{x-2}$.
Find each of the following:
(i) fof
(ii) fofof
(iii) (fofof) (38)
(iv) f2
Also, show that $f \circ f \neq f^{2}$.
$f(x)=\sqrt{x-2}$
For domain,
$x-2 \geq 0$
$\Rightarrow x \geq 2$
Domain of $f=[2, \infty)$
Since $f$ is a square-root function, range of $f=(0, \infty)$
So, $f:[2, \infty) \rightarrow(0, \infty)$
(i) fof
Range of $f$ is not a subset of the domain of $f$.
$\Rightarrow$ Domain $(f o f)=\{x: x \in$ domain of $f$ and $f(x) \in$ domain of $f\}$
$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \in[2, \infty)\}$
$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \geq 2\}$
$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $x-2 \geq 4\}$
$\Rightarrow$ Domain $($ fof $)=\{x: x \in[2, \infty)$ and $x \geq 6\}$
$\Rightarrow$ Domain $($ fof $)=\{x: x \geq 6\}$
$\Rightarrow$ Domain $($ fof $)=[6, \infty)$
$f o f:[6, \infty) \rightarrow R$
$(f o f)(x)=f(f(x))$
$=f(\sqrt{x-2})$
$=\sqrt{\sqrt{x-2}-2}$
(ii) fofof $=(f o f)$ of
We have, $f:[2, \infty) \rightarrow(0, \infty)$ and fof : $[6, \infty) \rightarrow R$
$\Rightarrow$ Range of $f$ is not a subset of the domain of $f o f$.
Then, domain $((f o f) o f)=\{x: x \in$ domain of $f$ and $f(x) \in$ domain of $f o f\}$
$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \in[6, \infty)\}$
$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $\sqrt{x-2} \geq 6\}$
$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $x-2 \geq 36\}$
$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \in[2, \infty)$ and $x \geq 38\}$
$\Rightarrow$ Domain $(($ fof $) o f)=\{x: x \geq 38\}$
$\Rightarrow$ Domain $(($ fof $) o f)=[38, \infty)$
fof $:[38, \infty) \rightarrow R$
So, $((f$ of $)$ of $)(x)=($ fof $)(f(x))$
$=(f$ of $)(\sqrt{x-2})$
$=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$
(iii) We have, (fofof) $(x)=\sqrt{\sqrt{\sqrt{x-2}-2}-2}$
So, $($ fofof $)(38)=\sqrt{\sqrt{\sqrt{38-2}-2}-2}$
$=\sqrt{\sqrt{\sqrt{36}-2}-2}$
$=\sqrt{\sqrt{6-2}-2}$
$=\sqrt{2-2}$
$=0$
(iv) We have, $f o f=\sqrt{\sqrt{x-2}-2}$ $\Rightarrow f^{2}(x)=f(x) \times f(x)=\sqrt{x-2} \times \sqrt{x-2}=x-2$
So, fof $\neq f^{2}$
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