## Calculus 8th Edition

(a) As per intermediate value theorem $e^{x}+x=0$ has one root at (-1,0) (b) Hence, the root of the equation $e^{x}+x=0$ is $x=-0.567143$
(a) Consider $f(x) =e^{x}+x=0$ Since $f(x) =e^{x}+x$ is continuous on all real numbers. Also $f(-1)=e^{-1}-1<0$ and $f(0)=e^{0}+0=1>0$ As per intermediate value theorem $e^{x}+x=0$ has one root at (-1,0) (b) Using Newton’s approximation formula. $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ Take first approximation $x_{1}=-0.5$ because equation $f(x) =e^{x}+x=0$ can be equal to zero if $x<0$. Also, $f'(x) =e^{x}+1$ Thus, $x_{2}=x_{1}-\frac{e^{x_{1}}+x_{1}}{e^{x_{1}}+1}$ $x_{2}=(-0.5)-\frac{e^{-0.5}+(-0.5)}{e^{(-0.5)}+1}$ $x_{2}\approx-0.566311$ $x_{3}=(-0.566311)-\frac{e^{-0.566311}+(-0.566311)}{e^{(-0.566311)}+1}$ $x_{3}\approx-0.567143$ and Proceeding in the same way $x_{4}\approx-0.567143$ Therefore, $x_{3}\approx x_{4}$. Hence, the root of the equation $e^{x}+x=0$ is $x=-0.567143$