Answer
(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 $x=-0.567143$
Work Step by Step
(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 $x=-0.567143$
