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

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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:
LHL=$\lim\limits_{\large x\to 2^-}(x^3-3)$
$\quad\quad=(8-3)$
$\quad\quad=5$
RHL=$\lim\limits_{\large x\to 2^+}(x^2+1)$
$\quad\quad=(4+1)$
$\quad\quad=5$
F(2)=$2^3-3$
$\;\;\;\;\;=8-3$
$\;\;\;\;\;=5$
$f$ is continuous at $x=2$
LHL=RHL.
Step 2:
At $x=c < 2$
$\lim\limits_{\large x\to c}(x^3-3)=(c^3-3)$
$\qquad\qquad\;\;\;=f(c)$
At $x=c > 2$
$\lim\limits_{\large x\to c}(x^2+1)=(c^2+1)$
$\qquad\qquad\;\;\;=f(c)$
Step 3:
$f$ is continuous for all $x\in R$
There is no point of discontinuity.
answered May 28, 2013 by sreemathi.v
 

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