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Find all points of discontinuity of \(f\), where \(f\) is defined by $f(x) = \left\{ \begin{array} {1 1} 2x + 3,& \quad\text{ if $ x $ \(\leq 2\)}\\ 2x - 3,& \quad \text{if $x$ > 1}\\ \end{array} \right. $

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

1 Answer

  • 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 $x=2$
LHL=$\lim\limits_{\large x\to 2}(2x+3)=7$
$f(2)=2\times 2+3=7$
RHL=$\lim\limits_{\large x\to 2}(2x-3)=7$
$\quad\;\;=2\times 2-3$
LHL$\neq$ RHL
$f$ is discontinuous at $x=2$
Step 2:
At $x=c<2\lim\limits_{\large x\to c}(2x+3)$
$f$ is continuous at $x=c<2$
Step 3:
At $x=c>2\lim\limits_{\large x\to c}(2x-3)=2c-3$
Therefore $f$ is continuous at $x=c$
Point of discontinuity is $x=2$.
answered May 27, 2013 by sreemathi.v

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