1. Tardigrade
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  4. The function f(x) = begincases x2 textfor x < 1 2 - x textfor x ≥ 1 endcases is

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Q. The function $f(x) = \begin{cases} x^2 & \quad \text{for } x < 1\\ 2 - x & \quad \text{for } x \geq 1 \end{cases}$ is

COMEDKCOMEDK 2007Continuity and Differentiability

Solution:

$f(x) = \begin{cases} x^2 & \quad \text{for } x < 1\\ 2 - x & \quad \text{for } x \geq 1 \end{cases}$
$\displaystyle\lim_{x\to1^{-}} f\left(x\right) =\displaystyle\lim _{x\to 1^{+}} f\left(x\right) =f\left(1\right)$
$\therefore \:\: \displaystyle\lim _{x\to 1^{-}} x^{2} =\displaystyle\lim _{x \rightarrow1^{+}} 2-x = 2-1 $
Hence, $f(x)$ is continuous at $x = 1$
Now, $ f'(x) = \begin{cases} 2x & \quad \text{for } x < 1\\ -1 & \quad \text{for } x \geq 1 \end{cases} $
$ \therefore \:\: \displaystyle\lim_{x\to1^{-}} f'\left(x\right) = \displaystyle\lim _{x\to 1^{-}} 2x =2$
$ \displaystyle\lim _{x\to 1^{+}} f'\left(x\right) = \displaystyle\lim _{x\to 1^{-}} 2x = 2 \displaystyle\lim _{x\to 1^{+} } f'\left(x\right)= \displaystyle\lim _{x\to 1^{+}} \left(-x\right)=-1 $
$\Rightarrow \displaystyle\lim _{x\to 1^{-}} f\left(x\right) \ne \displaystyle\lim _{x\to 1^{+} } f'\left(x\right) $
$i.e., L.H.D \neq R.H.D.$
Hence. $f(x)$ is.no] differetiable at $x = 1$