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If the function is $f(x)=\frac{1}{x+2}$, then the point of discontinuity of the composite function $y=f(f(x))$ is
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The correct answer is:
$\frac{-5}{2}$
$y=f(f(x))=f\left(\frac{1}{2+x}\right)$
$=\frac{1}{2+\frac{1}{2+x}}=\frac{2+x}{4+2 x+1}=\frac{2+x}{5+2 x}$
This function is not continuous at $x=-2$ (because $f(x)$ is not defined at this point) and $x=-\frac{5}{2}$ because $f(f(x))$ is not defined at this point.
$=\frac{1}{2+\frac{1}{2+x}}=\frac{2+x}{4+2 x+1}=\frac{2+x}{5+2 x}$
This function is not continuous at $x=-2$ (because $f(x)$ is not defined at this point) and $x=-\frac{5}{2}$ because $f(f(x))$ is not defined at this point.
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