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Find all the points of discontinuity of $f$ defined by $f (x) = | x | – | x + 1 |.$

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Toolbox:
  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$f(x)=|x|-|x+1|$
When $x < -1$
$f(x)=-x-[-(x+1)]$
$\quad\;\;\;=-x+x+1$
$\quad\;\;\;=1$
When $-1\leq x< 0$
$f(x)=-x-(x+1)$
$\quad\;\;\;=-x-x-1$
$\quad\;\;\;=-2x-1$
When $x\geq 0$
$f(x)=x-(x+1)$
$\quad\;\;\;=x-x-1$
$\quad\;\;\;=-1$
$f(x)=\left\{\begin{array}{1 1}1 &if\;x<-1\\-2x-1 & if\;-1\leq 0\\-1 &if\;x\geq 0\end{array}\right.$
Step 2:
At $x=-1$
LHL=$\lim\limits_{\large x\to 1^-}f(x)$
$\quad\;\;=\lim\limits_{\large x\to 1^-}(1)$
$\quad\;\;=1$
RHL=$\lim\limits_{\large x\to (-1)^+}f(x)$
$\quad\;\;=\lim\limits_{\large x\to (-1)^+}(-2x-1)$
$\quad\;\;\;=-2(-1)-1$
$\quad\;\;\;=2-1$
$\quad\;\;\;=1$
$f(-1)=-2(-1)-1$
$\quad\;\;\;\;\;\;=2-1$
$\quad\;\;\;\;\;\;=1$
LHL=RHL=$f(-1)$
$\Rightarrow f$ is continuous at $x=-1$
Step 3:
At $x=0$
LHL=$\lim\limits_{\large x\to 0^-}(-2x-1)$
$\qquad=(-2(0)-1)$
$\qquad=-1$
$f(0)=-1$
RHL=$\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large x\to 0^+}(-1)=-1$
LHL=RHL=$f(0)$
$f$ is continuous at $x=0$
Hence $f$ is continuous for all $x\in R$
answered May 30, 2013 by sreemathi.v
 

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