Question Number 168428 by mnjuly1970 last updated on 10/Apr/22
$$ \\ $$$$\:\:\:\:\:{solve} \\ $$$$\:\:\:\:\:{f}\left({x}\right)=\:{x}\:+\sqrt{{x}^{\:\mathrm{2}} −\mathrm{3}{x}\:+\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{R}\:_{{f}} \:=?\:\:\:\:\:\:\:{R}_{\:{f}} :\:{rang}\:{of}\:\:\:\:{f} \\ $$
Answered by MJS_new last updated on 11/Apr/22
$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\leqslant\mathrm{1}\:\mathrm{oe}\:{x}\geqslant\mathrm{2} \\ $$$$\mathrm{for}\:−\infty<{x}\leqslant\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{3}}{\mathrm{2}}>{f}\left({x}\right)\geqslant\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{2}\leqslant{x}<+\infty\:\mathrm{we}\:\mathrm{get}\:\mathrm{2}\leqslant{f}\left({x}\right)<+\infty \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{is}\:\left[\mathrm{1};\:\frac{\mathrm{3}}{\mathrm{2}}\right)\cup\left[\mathrm{2};\:+\infty\right) \\ $$