જો $f(x) = \left\{ \begin{array}{l}x + 2\,, - 1 < x < 3\\5\,\,\,\,\,\,\,\,,\,\,\,\,x = 3\\8 - x\,,\,\,\,\,x > 3\end{array} \right.$, તો $f'(x) $ એ $x = 3$ આગળ મેળવો.
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Answer
If $f(x) = \left\{ \begin{array}{l}x + 2\,,\,\,\, - 1 < x < 3\\\,\,\,\,\,\,5\,\,,\,\,\,\,\,\,\,\,\,\,\,\,x = 3\\8 - x\,\,,\,\,\,\,\,\,\,\,\,\,\,x > 3\end{array} \right.$and $f(3) = 5$
$\text{L.H.D} = \mathop {\lim }\limits_{x \to 3 - } \frac{{f(x) - f(3)}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} \frac{{f(3 - h) - f(3)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{(3 - h + 2) - 5}}{{ - h}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{ - h}}$
$= 1$
$\text{R.H.D} = \mathop {\lim }\limits_{x \to {3^ + }} \frac{{f(x) - f(3)}}{{x - 3}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{f(3 + h) - f(3)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{8 - (3 + h) - 5}}{h}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{h}$
$= - 1$
$\text{L.H.D} \ne \text{R.H.D}$
$f(x)$ is not differentiable.
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