(9am to 6pm)

Ask Questions, Get Answers

Want help in doing your homework? We will solve it for you. Click to know more.

Find all points of discontinuity of \(f\), where \(f\) is defined by $ f(x) = \left\{ \begin{array} {1 1} x^{10} -1 ,& \quad\text{ if $ x $ \(\leq 1\)}\\ x^2,& \quad \text{if $x$ > 1}\\ \end{array} \right. $

$\begin{array}{1 1} \text{Point of discontinuity is x=1} \\ \text{Point of discontinuity is x=-1} \\ \text{Points of discontinuity are all x<1} \\ \text{Points of discontinuity are all x > 1}\end{array} $

1 Answer

Need homework help? Click here.
  • 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 =$\lim\limits_{\large x\to 1^-}f(x)=(x^{10}-1)$
RHL =$\lim\limits_{\large x\to 1^+}f(x)=(x^2)$
$\Rightarrow $LHL $\neq$ RHL $\neq$ f(1)
$f$ is not continuous at $x=1$
Step 2:
At $x=c < 1$
$\lim\limits_{\large x\to c}(x^{10}-1)=(c^{10}-1)$
At $x=c > 1$
$\lim\limits_{\large x\to c}(x^2)=(c^2)$
Step 3:
$f$ is continuous at all points $x\in R-\{1\}$
Therefore point of discontinuity is $x=1$
answered May 28, 2013 by sreemathi.v
edited Jul 1, 2014 by rvidyagovindarajan_1

Related questions