Question Number 211637 by CrispyXYZ last updated on 15/Sep/24
$${f}\left({x}\right)\:=\:\mathrm{sin}\:{x}\:−\:\mathrm{e}^{{x}} \:+\:\mathrm{1}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)\:\mathrm{has}\:\mathrm{only}\:\mathrm{2}\:\mathrm{zeros}\:\mathrm{in}\:−\pi\leqslant{x}\leqslant\mathrm{0}. \\ $$
Answered by mehdee1342 last updated on 15/Sep/24
$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(−\pi\right)=\mathrm{1}−{e}^{−\pi} \mathrm{0}\:\:\:\&\:\:\:{f}\left(−\frac{\pi}{\mathrm{2}}\right)=−{e}^{−\frac{\pi}{\mathrm{2}}} <\mathrm{0} \\ $$$$\Rightarrow\exists\:\:−\pi<{a}<−\frac{\pi}{\mathrm{2}}\:\:;\:\:\:{f}\left({a}\right)=\mathrm{0} \\ $$$$ \\ $$