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Discuss the continuity of the function \(f\), where \(f\) is defined by $ f(x) = \left\{ \begin{array} {1 1} 2x ,& \quad\text{ if $ x $ < 0}\\ 0,& \quad\text{if $ 0 $ \(\leq x\) \(\leq1\)}\\ 4x ,& \quad\text{ if $ x $ > 1}\\ \end{array} \right. $

$\begin{array}{1 1} \text{The point of discontinuity is x=-1} \\ \text{The point of discontinuity is x=1} \\ \text{The point of discontinuity is x=0} \\ \text{The point of discontinuity is x=2}\end{array} $

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  • 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:
At $x=0$
LHL=$\lim\limits_{\large x\to 0}(2x)=0$
RHL=$\lim\limits_{\large x\to 0}(0)=0$
$f(x)=LHL=RHL=0$
$f$ is continuous at $x=0$
Step 2:
At $x=1$
LHL=$\lim\limits_{\large x\to 1}(0)=0$
RHL=$\lim\limits_{\large x\to 1^+}(4x)=4$
$f(1)=0$
$f(1)$=LHL $\neq$ RHL.
$f$ is not continuous at $x=1$
Step 3:
When $x < 0\; f(x)=2x$ being a polynomial it is discontinuous at point $x < 0$
When $x > 1\; f(x)=4x$ being a polynomial it is continuous at all point $x > 1$
When $0 \leq x\leq 1\;f(x)=0$ is a continuous function.
The point of discontinuity is $x=1$
answered May 28, 2013 by sreemathi.v
 

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