logo

Ask Questions, Get Answers

X
 

At the point $x=1$ the function $f(x)=\left\{\begin{array}{1 1}x^3-1&1 < x < \infty\\x-1&-\infty < x \leq 1\end{array}\right.$ is

$\begin{array}{1 1}(a)\;\text{Continuous and differentiable}\\(b)\;\text{Continuous and not differentiable}\\(c)\;\text{DisContinuous and differentiable}\\(d)\;\text{DisContinuous and not differentiable}\end{array}$

Download clay6 mobile app

1 Answer

$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 1}(x-1)=0$
$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 1}(x^3-1)=0$
also $f(1)=1-1=0$
$\therefore f$ is continuous at $x=1$
Clearly $Lf'(1)=2$ and $Rf'(1)=3$
$\therefore f(x)$ is not differentiable at $x=1$
Hence (c) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

Related questions

...