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Question Number 89834 by Cheyboy last updated on 19/Apr/20

Find x   e^x = x^2 −1  anyother method apart from Newton′s

$${Find}\:{x}\: \\ $$$$\boldsymbol{{e}}^{\boldsymbol{{x}}} =\:\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1} \\ $$$$\boldsymbol{{anyother}}\:\boldsymbol{{method}}\:\boldsymbol{{apart}}\:\boldsymbol{{from}}\:\boldsymbol{{N}}{ewton}'{s} \\ $$

Commented by Joel578 last updated on 19/Apr/20

bisection method, secant method

$${bisection}\:{method},\:{secant}\:{method} \\ $$

Answered by TANMAY PANACEA. last updated on 19/Apr/20

f(x)=e^x −x^2 +1  f(0)=2>0  f(1)=e>0  f(2)=e^2 −3>0  so no root when x>0  f(−1)=(1/e)>0  f(−2)=(1/e^2 )−3<0  so one root lies between   −1>x>−2

$${f}\left({x}\right)={e}^{{x}} −{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2}>\mathrm{0} \\ $$$${f}\left(\mathrm{1}\right)={e}>\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)={e}^{\mathrm{2}} −\mathrm{3}>\mathrm{0} \\ $$$${so}\:{no}\:{root}\:{when}\:{x}>\mathrm{0} \\ $$$${f}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{{e}}>\mathrm{0} \\ $$$${f}\left(−\mathrm{2}\right)=\frac{\mathrm{1}}{{e}^{\mathrm{2}} }−\mathrm{3}<\mathrm{0} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{one}}\:\boldsymbol{{root}}\:\boldsymbol{{lies}}\:\boldsymbol{{between}}\:\:\:−\mathrm{1}>\boldsymbol{{x}}>−\mathrm{2} \\ $$

Commented by Cheyboy last updated on 19/Apr/20

Alright thank sir

$${Alright}\:{thank}\:{sir} \\ $$

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