Question Number 89834 by Cheyboy last updated on 19/Apr/20
$${Find}\:{x}\: \\ $$$$\boldsymbol{{e}}^{\boldsymbol{{x}}} =\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1} \\ $$$$\boldsymbol{{anyother}}\:\boldsymbol{{method}}\:\boldsymbol{{apart}}\:\boldsymbol{{from}}\:\boldsymbol{{N}}{ewton}'{s} \\ $$
Commented by Joel578 last updated on 19/Apr/20
$${bisection}\:{method},\:{secant}\:{method} \\ $$
Answered by TANMAY PANACEA. last updated on 19/Apr/20
$${f}\left({x}\right)={e}^{{x}} −{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2}>\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)={e}>\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)={e}^{\mathrm{2}} −\mathrm{3}>\mathrm{0} \\ $$$${so}\:{no}\:{root}\:{when}\:{x}>\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{{e}}>\mathrm{0} \\ $$$${f}\left(−\mathrm{2}\right)=\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{3}<\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{one}}\:\boldsymbol{{root}}\:\boldsymbol{{lies}}\:\boldsymbol{{between}}\:\:\:−\mathrm{1}>\boldsymbol{{x}}>−\mathrm{2} \\ $$
Commented by Cheyboy last updated on 19/Apr/20
$${Alright}\:{thank}\:{sir} \\ $$