Question Number 122393 by benjo_mathlover last updated on 16/Nov/20
$$\:\:{f}\left({x}\right)\:=\:{x}\:\mid{x}−\mathrm{1}\mid\:\Rightarrow{f}\:'\left(\mathrm{1}\right)=? \\ $$
Answered by MJS_new last updated on 16/Nov/20
$$\mathrm{does}\:\mathrm{not}\:\mathrm{exist} \\ $$
Answered by Bird last updated on 17/Nov/20
$${lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\:\frac{{f}\left(\mathrm{1}+{h}\right)−{f}\left(\mathrm{1}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow{o}^{+} } \:\:\:\frac{\left(\mathrm{1}+{h}\right){h}}{{h}}\:={lim}_{{h}\rightarrow\mathrm{0}^{+} } \:\mathrm{1}+{h}=\mathrm{1} \\ $$$${lim}_{{h}\rightarrow\mathrm{0}^{−} } \:\:\:\frac{{f}\left(\mathrm{1}+{h}\right)−{f}\left(\mathrm{1}\right)}{{h}} \\ $$$$={lim}_{{h}\rightarrow\mathrm{0}^{−} } \frac{\left(\mathrm{1}+{h}\right)\left(−{h}\right)}{{h}}={lim}_{{h}\rightarrow\mathrm{0}^{−} } −\left(\mathrm{1}+{h}\right) \\ $$$$=−\mathrm{1}\:{its}\:{clear}\:{thst}\:{f}\:{is}\:{not}\:{derivable} \\ $$$${at}\:\mathrm{1}\:\Rightarrow{f}^{'} \left(\mathrm{1}\right)\:{dont}\:\:{exist} \\ $$