Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Given the function $ f(x)=\large\frac{1}{x+2}.$Find the points of discontinuity of the composite function $y=f(f(x)).$

$\begin{array}{1 1} x=-2,\large\frac{-5}{2} \\ x=2,\large\frac{5}{2} \\ x=-2,\large\frac{5}{2} \\ x=2,\large\frac{-5}{2} \end{array} $

Can you answer this question?

1 Answer

0 votes
  • A function is said to be discontinuous at a point if LHL $\neq$ RHL.
  • It is also discontinuous if both LHL and RHL does not exist.
Step 1:
Given function $f(x)=\large\frac{1}{x+2}$
Therefore $f(f(x))=\Large\frac{1}{\Large\frac{1}{x+2}\normalsize +2}$
On simplifying we get,
Step 2:
Now let us discuss the continuity of the function.
When $x=\large\frac{-5}{2}$
The function becomes undefined.
Hence clearly $x=-\large\frac{-5}{2}$ is one point of discontinuity .
Step 3:
When $x=-2$
The LHL :
$\lim\limits_{\large x\to -2^-}f(x)=\lim\limits_{\large x\to -2^-}\large\frac{x+2}{2x+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-2-h+2}{2(-2-h)+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-h}{(1-h)}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-1}{\Large\frac{1}{h}\normalsize -1}$
Step 4:
The RHL :
$\lim\limits_{\large x\to -2^+}f(x)=\lim\limits_{\large x\to -2^+}\large\frac{x+2}{2x+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-2+h+2}{2(-2+h)+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{h}{(1+h)}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{1}{\Large\frac{1}{h}\normalsize +1}$
Hence LHL $\neq$ RHL.
Hence the function is discontinuous at $x=-2$.
Hence the point of discontinuity are $x=-2$ and $\large\frac{-5}{2}$
answered Jun 26, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App