$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {{{(1 + h)}^3} - 1} \right\} - 0}}{h} = 3$
$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f(1 - h) - f(1)}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\left\{ {(1 - h) - 1} \right\} - 0}}{{ - h}} = 1$
$\therefore \,\,\,Rf'(1) \ne Lf'(1)$ $ \Rightarrow \,\,f(x)$ is not differentiable at $x = 1.$
Now, $f(1 + 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 + h) = 0$
and $f(1 - 0) = \mathop {\lim }\limits_{h \to 0} \,f(1 - h) = 0$
$\therefore \,\,\,f(1 + 0) = f(1 - 0) = f(0)$
$ \Rightarrow \,\,f(x)$ is continuous at $x = 1.$
Hence at $x = 1,\,\,\,f(x)$ is continuous and not differentiable.
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Statement $-2:$ The functions $x^2e^x$ and $x^2e^{-x}$ are increasing for all $x > 0$ and the sum of two increasing functions in any interval $(a, b)$ is an increasing function in $(a, b).$