Question Number 41101 by ajfour last updated on 02/Aug/18
$${Find}\:{a}\:{general}\:{formula}\:{for}\:{m} \\ $$$${such}\:{that}\:{f}\left({x}\right)=\mid\mathrm{sin}\:{x}\mid+\mathrm{sin}\:\mid{x}\mid \\ $$$${be}\:{not}\:{differentiable}\:{at}\:{x}={m}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Aug/18
Commented by ajfour last updated on 02/Aug/18
$${good}\:{sir};\:{thanks}. \\ $$
Answered by MrW3 last updated on 02/Aug/18
$${f}\left(−{x}\right)={f}\left({x}\right)\:\Rightarrow\:{we}\:{only}\:{need}\:{to}\:{see}\:{x}\geqslant\mathrm{0}: \\ $$$${f}\left({x}\right)=\mid\mathrm{sin}\:{x}\mid+\mathrm{sin}\:{x} \\ $$$$\mid\mathrm{sin}\:{x}\mid=\mathrm{sin}\:{x}\:{for}\:\mathrm{2}{n}\pi\leqslant{x}\leqslant\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\:{with}\:{n}\geqslant\mathrm{0} \\ $$$$\mid\mathrm{sin}\:{x}\mid=−\mathrm{sin}\:{x}\:{for}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\leqslant{x}\leqslant\left(\mathrm{2}{n}+\mathrm{2}\right)\pi \\ $$$$ \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2sin}\:{x}\:{for}\:\mathrm{2}{n}\pi\leqslant{x}\leqslant\left(\mathrm{2}{n}+\mathrm{1}\right)\pi \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{0}\:{for}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\leqslant{x}\leqslant\left(\mathrm{2}{n}+\mathrm{2}\right)\pi \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{not}\:{differentiable}\:{at}\:{x}=\mathrm{2}{n}\pi\:{and}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi, \\ $$$${so}\:{generally} \\ $$$$\Rightarrow{m}={k}\pi\:{with}\:{k}={any}\:{integer} \\ $$
Commented by ajfour last updated on 02/Aug/18
$${Absolutely}\:{true},\:{Sir}.\:{I}\:{had}\:{drawn} \\ $$$${the}\:{graph}. \\ $$