Question Number 57406 by Abdo msup. last updated on 03/Apr/19
$${let}\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{cos}\left(\pi{x}\right) \\ $$$$\left.\mathrm{1}\left.\right)\:{prove}\:{that}\:{f}\left({x}\right)=\mathrm{0}\:{have}\:{a}\:{solurion}\:\alpha\:{inside}\:\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\left.\mathrm{2}\right)\:{use}\:{newton}\:{method}\:{to}\:{find}\:{a}\:{approximate}\: \\ $$$${value}\:{of}\:\alpha\:. \\ $$
Commented by kaivan.ahmadi last updated on 04/Apr/19
$${f}\left(\mathrm{0}\right)=\mathrm{1}>\mathrm{0}\:\:,{f}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}<\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{0}\right){f}\left(\mathrm{1}\right)<\mathrm{0}\Rightarrow{f}\:{has}\:{at}\:{least}\:{one}\: \\ $$$${slution}\:{in}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$