Question Number 206150 by Samuel12 last updated on 08/Apr/24
$$\mathrm{Calcul}\:\:\:\mid\mathrm{x}−\alpha\mid\:=\:????? \\ $$$$\:\:\bullet\:\:\:\:\:\:\mathrm{x}=−\mathrm{1}\:\:;\:\:\:\alpha\:\in\:\left[−\mathrm{1};\:−\frac{\mathrm{3}}{\mathrm{4}}\right] \\ $$$$\:\:\bullet\:\:\:\:\mathrm{x}=\mathrm{2}\:\:;\:\:\:\:\alpha\:\in\:\left[\mathrm{0}\:;\:\mathrm{3}\right]\:\:\:\:\:\:\:\mathrm{help}\:\mathrm{please}\:\: \\ $$
Answered by Frix last updated on 08/Apr/24
$$\mid{x}−\alpha\mid=\begin{cases}{{x}−\alpha;\:{x}−\alpha\geqslant\mathrm{0}}\\{−\left({x}−\alpha\right);\:{x}−\alpha<\mathrm{0}}\end{cases} \\ $$$$\bullet\:{x}=−\mathrm{1}\wedge−\mathrm{1}\leqslant\alpha\leqslant−\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow\:−\frac{\mathrm{1}}{\mathrm{4}}\leqslant{x}−\alpha\leqslant\mathrm{0} \\ $$$$\Rightarrow\:\mid{x}−\alpha\mid=−\left({x}−\alpha\right) \\ $$$$\bullet\:{x}=\mathrm{2}\wedge\mathrm{0}\leqslant\alpha\leqslant\mathrm{3}\:\Rightarrow\:−\mathrm{1}\leqslant{x}−\alpha\leqslant\mathrm{2} \\ $$$$\Rightarrow\:\mid{x}−\alpha\mid=\begin{cases}{{x}−\alpha;\:\mathrm{0}\leqslant\alpha\leqslant\mathrm{2}}\\{−\left({x}−\alpha\right);\:\mathrm{2}<\alpha\leqslant\mathrm{3}}\end{cases} \\ $$