Question Number 70741 by aliesam last updated on 07/Oct/19
$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{}\\{\mid{x}−\mathrm{2}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{1}}\end{cases} \\ $$ $$ \\ $$ $${find}\:{the}\:{critical}\:{points} \\ $$
Commented bykaivan.ahmadi last updated on 07/Oct/19
$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{\mathrm{2}−{x}\:\:\:\:\:\:\mathrm{1}<{x}<\mathrm{2}}\\{{x}−\mathrm{2}\:\:\:\:\:\:\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$ $${f}'\left({x}\right)=\begin{cases}{\mathrm{2}{x}\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}<{x}<\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{2}}\end{cases} \\ $$ $${f}'\left({x}\right)=\mathrm{0}\Rightarrow\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\mathrm{0}\:\:{is}\:{a}\:{critical}\:{point} \\ $$ $${f}'\left(\mathrm{1}\right),{f}'\left(\mathrm{2}\right)\:{are}\:{not}\:{exist}\:{so}\:{they}\:{are}\:{critical}\:{points} \\ $$ $$ \\ $$
Commented byaliesam last updated on 07/Oct/19
$${f}'\left(\mathrm{1}\right)=? \\ $$
Commented byaliesam last updated on 07/Oct/19
Commented bykaivan.ahmadi last updated on 07/Oct/19
$${you}\:{are}\:{right} \\ $$
Commented byaliesam last updated on 07/Oct/19
$${ok}\:{sir}\:{and}\:{thank}\:{you}\:{for}\:{the}\:{solution}\: \\ $$