Question Number 75083 by Rio Michael last updated on 07/Dec/19 | ||
$${A}\:{function}\:{f}\:{is}\:{given}\:{by}\: \\ $$ $$\:{f}\left({x}\right)\:=\:\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3},\:\:\mathrm{0}\leqslant{x}<\mathrm{2}}\\{\mathrm{4}{x}−\mathrm{7},\:\mathrm{2}\leqslant{x}<\mathrm{4}}\end{cases} \\ $$ $${is}\:{such}\:{that}\:{f}\left({x}\right)\:=\:{f}\left({x}\:+\:\mathrm{4}\right)\: \\ $$ $${find}\:\:{f}\left(\mathrm{27}\right)\:{and}\:{f}\left(−\mathrm{106}\right). \\ $$ | ||
Answered by MJS last updated on 07/Dec/19 | ||
$$\mathrm{27}=\mathrm{6}×\mathrm{4}+\mathrm{3} \\ $$ $$−\mathrm{106}=−\mathrm{27}×\mathrm{4}+\mathrm{2} \\ $$ $$\Rightarrow \\ $$ $${f}\left(\mathrm{27}\right)={f}\left(\mathrm{3}\right)=\mathrm{5} \\ $$ $${f}\left(−\mathrm{106}\right)={f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$ | ||
Commented byRio Michael last updated on 07/Dec/19 | ||
$${thank}\:{you}\:{sir},\:{i}\:{got}\:{the}\:{now} \\ $$ | ||