$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 - h) = \mathop {\lim }\limits_{h \to 0} (1 - h) = 1$
$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(1 + h) = \mathop {\lim 2}\limits_{h \to 0} (1 + h) - 1 = 1$
$\because \,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=1$
$\therefore$ Function is continuous at $x = 1$.
$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 - h) - f(1)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{(1 - h) - 1}}{{ - h}} = 1$
$Rf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{2 + 2h - 1 - 1}}{h} = 2$
$\therefore$ $Lf'(1) \ne Rf'(1)$
$\therefore$ Function is not differentiation at $x = 1$
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$\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is