Question:
If $f(x) = \begin{cases} xe^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}, & \text{$x \ne0$} \\[2ex] 0, & \text{$x=0$} \end{cases}$ then $f (x)$ is
If $f(x) = \begin{cases} xe^{-\left(\frac{1}{\left|x\right|}+\frac{1}{x}\right)}, & \text{$x \ne0$} \\[2ex] 0, & \text{$x=0$} \end{cases}$ then $f (x)$ is
Updated On: Jul 27, 2022
- continuous as well as differentiable for all $x$
- continuous for all x but not differentiable at $x = 0$
- neither differentiable nor continuous at $x = 0$
- discontinuous everywhere
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The Correct Option is B
Solution and Explanation
$f' (0)$
$f' (0 - h) = 1$
$f' (0 + h) = 0$
$LHD ? RHD.$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.