Answer:
\[Rf'(2)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2+h)-f(2)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{5-(2+h)-3}{h} \right]\] \[=\underset{h\to 0}{\mathop{lim}}\,\frac{-\,h}{h}=\underset{h\to 0}{\mathop{lim}}\,(-\,1)=-\,1\] and\[Lf'(2)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2-h)-f(2)}{-\,h}=\underset{h\to 0}{\mathop{\lim }}\,\left[ \frac{1+(2-h)-3}{-\,h} \right]\] \[=\underset{h\to 0}{\mathop{lim}}\,\frac{-\,h}{-\,h}=\underset{h\to 0}{\mathop{lim}}\,1=1\] Thus, \[Rf'(2)\ne Lf'(2).\] Hence, f(x) is not differentiable at x = 2.
You need to login to perform this action.
You will be redirected in
3 sec