f-x-x-2-x-1-x-2-x-gt-1-find-the-critical-points-

Question Number 70741 by aliesam last updated on 07/Oct/19

f(x)= { ((x^2 x≤1)),(),((∣x−2∣ x>1)) :} find the critical points

$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{}\\{\mid{x}−\mathrm{2}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{1}}\end{cases} \\ $$$$ \\ $$$${find}\:{the}\:{critical}\:{points} \\ $$

Commented by kaivan.ahmadi last updated on 07/Oct/19

f(x)= { ((x^2 x≤1)),((2−x 1<x<2)),((x−2 x≥2)) :} f′(x)= { ((2x x≤1)),((−1 1<x<2)),((1 x>2)) :} f′(x)=0⇒2x=0⇒x=0 is a critical point f′(1),f′(2) are not exist so they are critical points

$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{\mathrm{2}−{x}\:\:\:\:\:\:\mathrm{1}<{x}<\mathrm{2}}\\{{x}−\mathrm{2}\:\:\:\:\:\:\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${f}'\left({x}\right)=\begin{cases}{\mathrm{2}{x}\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}<{x}<\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{2}}\end{cases} \\ $$$${f}'\left({x}\right)=\mathrm{0}\Rightarrow\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\mathrm{0}\:\:{is}\:{a}\:{critical}\:{point} \\ $$$${f}'\left(\mathrm{1}\right),{f}'\left(\mathrm{2}\right)\:{are}\:{not}\:{exist}\:{so}\:{they}\:{are}\:{critical}\:{points} \\ $$$$ \\ $$

Commented by aliesam last updated on 07/Oct/19