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If the function $f(x)=\frac{1}{x+2}$ then find the points of discontinuity of the composite function $y=f\{f(x)\}$.
Solution:
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Verified Answer
We have, $f(x)=\frac{1}{x+2}$
$$
\begin{aligned}
&\therefore \mathrm{y}=\mathrm{f}\{\mathrm{f} x)\} \\
&\Rightarrow \mathrm{y}=\mathrm{f}\left(\frac{1}{\mathrm{x}+2}\right)=\frac{1}{\frac{1}{\mathrm{x}+2}+2} \Rightarrow \mathrm{y}=\frac{(\mathrm{x}+2)}{(2 \mathrm{x}+5)}
\end{aligned}
$$
Here, the function $y$ will not be continuous at those points, where it is not defined as it is a rational function. Therefore, $y=\frac{x+2}{(2 x+5)}$ is not defined, when $2 x+5=0$
$$
\therefore \mathrm{x}=\frac{-5}{2}
$$
Hence, $y$ is discontinuous at $x=\frac{-5}{2}$.
$$
\begin{aligned}
&\therefore \mathrm{y}=\mathrm{f}\{\mathrm{f} x)\} \\
&\Rightarrow \mathrm{y}=\mathrm{f}\left(\frac{1}{\mathrm{x}+2}\right)=\frac{1}{\frac{1}{\mathrm{x}+2}+2} \Rightarrow \mathrm{y}=\frac{(\mathrm{x}+2)}{(2 \mathrm{x}+5)}
\end{aligned}
$$
Here, the function $y$ will not be continuous at those points, where it is not defined as it is a rational function. Therefore, $y=\frac{x+2}{(2 x+5)}$ is not defined, when $2 x+5=0$
$$
\therefore \mathrm{x}=\frac{-5}{2}
$$
Hence, $y$ is discontinuous at $x=\frac{-5}{2}$.
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