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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} $

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Toolbox:
  • 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,
$\qquad\qquad\qquad=\large\frac{1}{\Large\frac{1+2(x+2)}{x+2}}$
$f(f(x))=\large\frac{x+2}{2x+5}$
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 :
$\lim\limits_{\large x\to -2^-}f(x)=\lim\limits_{\large x\to -2^-}\large\frac{x+2}{2x+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-2-h+2}{2(-2-h)+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-h}{(1-h)}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-1}{\Large\frac{1}{h}\normalsize -1}$
$\qquad\qquad\;\;=-1$
Step 4:
The RHL :
$\lim\limits_{\large x\to -2^+}f(x)=\lim\limits_{\large x\to -2^+}\large\frac{x+2}{2x+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{-2+h+2}{2(-2+h)+5}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{h}{(1+h)}$
$\qquad\qquad\;\;=\lim\limits_{\large x\to -2}\large\frac{1}{\Large\frac{1}{h}\normalsize +1}$
$\qquad\qquad\;\;=1$
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
 

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