Solving the equation: $ (x - 1)e^x + x + 2 = 0$

Question

We have : $(x - 1)e^x + x + 2 = 0$.

From the equation we have: $e^x = \frac{x + 2}{1 - x} \implies $ $x \in (-2,1)$.

I don't know how to proceed to isolate $x$.

Answer 1

Good news

Rewriting the equation as $$e^{-x} =- \frac{x-1 }{x + 2} $$ there is an explicit solution in terms of the generalized Lambert function (have a look at equation $(4)$).

Bad news : This is nice but not useful at all.

So, consider that you look for the zero of function $$f(x)=(x - 1)e^x + x + 2 \qquad \text{for} \qquad x \in (-2,1)$$ By inspection, you can see that the solution is between $-2$ (since $f(-2)=-\frac{3}{e^2}<0$) and $-1$ (since $f(-1)=1-\frac{2}{e}>0$).

So, use Newton method at the midpoint that is to say $x_0=-\frac 32$. The successive iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & -1.500000000 \\ 1 & -1.413084343 \\ 2 & -1.412461889 \\ 3 & -1.412461859 \end{array} \right)$$ which is quite fast.

Just for you curiosity, using the simplest $[1,1]$ Padé approximant built around $x_0=-\frac 32$, we could have, as an approximation, $$x \sim -\frac{99-121 e^{3/2}+32 e^3}{2 \left(13-23 e^{3/2}+8 e^3\right)} \approx -1.412446300$$

Using the $[1,2]$ Padé approximant would give $$x \sim \frac{3 \left(177-492 e^{3/2}+329 e^3-64 e^{9/2}\right)}{2 \left(-47+236 e^{3/2}-203 e^3+48 e^{9/2}\right)}\approx -1.412460337$$

Answer 2

You are not supposed to solve for $x$ here, since the problem is about the range of $x$, it is very likely that you will not be able to get an explicit expression for $x$ (at your level of knowledge, at least).

We have $f(x) = (x - 1)e^x + x + 2 = 0$. $f(1) = 3 > 0$, and since $(x-1)$ and $e^x$ and $x$ and $2$ are all positive increasing functions, $f(x)$ is increasing. So $f$ will only get bigger and bigger. And there will be no root past $x=1$. Formally, $f(x) = (x - 1)e^x + x + 2 > (1 - 1) e^1 + 1 + 2 =3 > 0$. On the other side, if $x \le 2$, $(x-1)e^x$ becomes negative, and $f(x) < x+2 \le 0$. So there is no root there either. This completes to proof.