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If we have: $f(x)=\frac { 1+x }{ 1+{ e }^{ x } } $

I am told to determine if $f(x)=x$ has multiple roots on $\left[ 0;+\infty \right] $

I tried to manually solve this equation, but I don't understand the result:

$f(x)-x=0\rightarrow \frac { 1+x }{ 1+{ e }^{ x } } =0\rightarrow { e }^{ x }(x+1)=0$

which would mean that -1 is the only solution. So there shouldn't be any solution on $\left[ 0;+\infty \right] $.

But when I check on Wolfram, I see that there is a unique solution, and that this solution is 0.56.

Can anybody help please ?

1) How can I find the nb of solutions to this equation ? 2) How can we explain the result I get and the one from wolfram ?

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If $f(x)=x$, then $f(x)-x=0$, or: $$\frac{1-xe^x}{1+e^x}=0.\tag{1}$$ Since $1+e^x$ is always positive, $(1)$ is equivalent to: $$ g(x)=xe^x = 1.\tag{2}$$ $g(x)$ is a decreasing and negative function over $(-\infty,-1)$, an increasing function over $(-1,+\infty)$, since $g'(x) = (x+1)e^x$. So there is at most one real number $x$ such that $xe^x=1$. Exactly one since $xe^x$ over $\mathbb{R}^+$ is positive and unbounded. Such a number is just: $$ W(1) = 0.567143290409783873\ldots $$ where $W(\cdot)$ is the Lambert W-function. $W(1)$ can be computed through Newton's method.

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  • $\begingroup$ thank you. But, was I supposed to know that the solution was W(1). Or can we manually solve this? $\endgroup$ – XCoder Sep 6 '14 at 22:55
  • $\begingroup$ Once you know that there is only one solution, you can find it numerically through Newton's method. $\endgroup$ – Jack D'Aurizio Sep 6 '14 at 22:57
  • $\begingroup$ How did you find out the variations please ? You did not use any derivative or anything. and how did you find out the -1 ? $\endgroup$ – XCoder Sep 6 '14 at 22:58
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    $\begingroup$ We can accomplish to this task by studying the simpler ausiliary function $g(x)-1$, whose set of zeroes is the same as $f(x)$. $\endgroup$ – Jack D'Aurizio Sep 6 '14 at 23:11
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    $\begingroup$ Thanks for your help and support Jack. Regards, $\endgroup$ – XCoder Sep 6 '14 at 23:15
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$$f(x) - x = \frac{1+x}{1+e^x} - x = \frac{1+x}{1+e^x} - \frac{x(1+e^x)}{1+e^x} = \frac{1-xe^x}{1+e^x}$$

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