Question Number 168114 by Beginner last updated on 03/Apr/22
Answered by MJS_new last updated on 04/Apr/22
$$\mathrm{7} \\ $$
Commented by Beginner last updated on 04/Apr/22
$${how}? \\ $$
Answered by MJS_new last updated on 04/Apr/22
$${f}_{\mathrm{1}} \left({x}\right)=\mid\mid{x}\mid−\mathrm{1}\mid=\begin{cases}{−\mathrm{2}<{x}<−\mathrm{1};\:−{x}−\mathrm{1}}\\{−\mathrm{1}\leqslant{x}<\mathrm{0};\:{x}+\mathrm{1}}\\{\mathrm{0}\leqslant{x}<\mathrm{1};\:\mathrm{1}−{x}}\\{\mathrm{1}\leqslant{x}<\mathrm{2};\:{x}−\mathrm{1}}\end{cases} \\ $$$${f}_{\mathrm{2}} \left({x}\right)=\mid\mathrm{cos}\:\pi{x}\mid=\begin{cases}{−\mathrm{2}<{x}\leqslant−\frac{\mathrm{3}}{\mathrm{2}};\:\mathrm{cos}\:\pi{x}}\\{−\frac{\mathrm{3}}{\mathrm{2}}<{x}\leqslant−\frac{\mathrm{1}}{\mathrm{2}};\:−\mathrm{cos}\:\pi{x}}\\{−\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}};\:\mathrm{cos}\:\pi{x}}\\{\frac{\mathrm{1}}{\mathrm{2}}<{x}\leqslant\frac{\mathrm{3}}{\mathrm{2}};\:−\mathrm{cos}\:\pi{x}}\\{\frac{\mathrm{3}}{\mathrm{2}}<{x}<\mathrm{2}:\:\mathrm{cos}\:\pi{x}}\end{cases} \\ $$$$\Rightarrow \\ $$$${f}_{\mathrm{1}} \left({x}\right)\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:{x}\in\left\{−\mathrm{1};\:\mathrm{9};\:\mathrm{1}\right\} \\ $$$${f}_{\mathrm{2}} \left({x}\right)\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:{x}\in\left\{−\frac{\mathrm{3}}{\mathrm{2}};\:−\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{3}}{\mathrm{2}}\right\} \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)\:\mathrm{not}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{7}\:\mathrm{points}\:\mathrm{within}\:\left(−\mathrm{2};\:\mathrm{2}\right) \\ $$
Commented by Beginner last updated on 04/Apr/22
$${Nice}\:{approach}\:{thx} \\ $$