Question Number 13955 by ajfour last updated on 25/May/17
$${y}\left({x}\right)=\begin{cases}{\mathrm{4}+\mathrm{6}{x}−\mathrm{3}{x}^{\mathrm{2}} \:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{20}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$ $${Is}\:{the}\:{function}\:{y}\left({x}\right)\: \\ $$ $${differentiable}\:{with}\:{respect}\:{to}\:{x} \\ $$ $${at}\:{x}=\mathrm{2}\:? \\ $$
Commented byprakash jain last updated on 25/May/17
$$\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:{y}\left({x}\right)=\mathrm{4}+\mathrm{12}−\mathrm{12}=\mathrm{4} \\ $$ $$\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\:{y}\left({x}\right)=\mathrm{8}−\mathrm{28}+\mathrm{20}=\mathrm{0} \\ $$ $${y}\left({x}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{continous}\:\mathrm{at}\:{x}=\mathrm{2}. \\ $$ $$\mathrm{Hence}\:\mathrm{not}\:\mathrm{differentiable}. \\ $$
Commented byajfour last updated on 25/May/17
$${very}\:{true},\:{thanks}\:{Sir}\:. \\ $$