If the function is f(x)=frac{1}{x+2},then the | vedclass

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(A) Given the function $f(x) = \frac{1}{x+2}$.For the composite function $y = f(f(x))$,we first note that $f(x)$ is undefined at $x = -2$.Now,calculat

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$-\frac{5}{2}$

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(A) Given the function $f(x) = \frac{1}{x+2}$.For the composite function $y = f(f(x))$,we first note that $f(x)$ is undefined at $x = -2$.Now,calculate $f(f(x)) = f\left(\frac{1}{x+2}ight) = \frac{1}{\frac{1}{x+2} + 2}$.Simplify the denominator: $\frac{1}{x+2} + 2 = \frac{1 + 2(x+2)}{x+2} = \frac{1 + 2x + 4}{x+2} = \frac{2x + 5}{x+2}$.Thus,$f(f(x)) = \frac{x+2}{2x+5}$.The composite function is undefined when the denominator $2x+5 = 0$,which gives $x = -\frac{5}{2}$.Also,the original function $f(x)$ must be defined,so $x 
eq -2$.Therefore,the points of discontinuity are $x = -2$ and $x = -\frac{5}{2}$.Comparing with the given options,the correct point of discontinuity is $-\frac{5}{2}$.

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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