question_answer

If the function \[f(x)=\frac{1}{x+2},\] find the points of discontinuity of the composite function \[y=f(f(x)).\]

Answer:

Given,   \[f(x)=\frac{1}{x+2}.\] Clearly, f(x) is not continuous at \[x=-\,2.\] [\[\therefore \] rational functions are continuous for all real numbers except at those points, where the denominator is zero] Now, for \[x\ne -\,2,\] \[f(f(x))=\frac{1}{f(x)+2}\]             \[=\frac{1}{\frac{1}{(x+2)}+2}=\frac{(x+2)}{1+2(x+2)}=\frac{x+2}{2x+5}\] which is discontinuous at \[x=-\frac{5}{2}.\]                   Hence, the points of discontinuity are                \[x=-\,2\] and \[x=-\frac{5}{2}.\]