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As the title mentioned, how to solve $x$ from the equation:

$$(x-1)e^{-x} > 0.5$$

How can I solve this analytically? This is a part of my homework and I got stuck to this equation. I'm also given the information that:

$$-x e^{-x} \leq -0.1x$$

if it helps. The problem is to find the bounds for $x$ in Wolfe's test.

P.S. do I need to use for example Newton's method to solve this?

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Hint

Consider the function $$f(x)=(x-1)e^{-x}-\frac12$$ Its erivative simplifies to $$f'(x)=-e^{-x} (x-2)$$ which means that there is an extremum at $x=2$ and $f(2)=\frac{1}{e^2}-\frac{1}{2}\approx -0.364665$. The second derivative test shows that this is a maximum.

I am sure that you can take from here.

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  • $\begingroup$ Thank you for your help =) $\endgroup$ – jjepsuomi Nov 12 '14 at 8:50
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    $\begingroup$ You are very welcome ! $\endgroup$ – Claude Leibovici Nov 12 '14 at 8:50
  • $\begingroup$ Why not consider the (simpler) function $f(x)=e^x-2(x-1)$ (and use the same crucial step that $e^2\gt2$)? $\endgroup$ – Did Nov 12 '14 at 8:51
  • $\begingroup$ @Did. "In mathematics house are many mansions" as almost said John (14:2) ! Cheers. I agree, your way is better. $\endgroup$ – Claude Leibovici Nov 12 '14 at 8:55
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Your inequality can be written as

$(1-x)e^{1-x}<-0.5e$

Then we need to solve $ye^{y} < -0.5e$ with $y=1-x$.

$g(y)=ye^y$, then $g'(y) = e^y + ye^y = e^y(1+y)$, so $g(y)$ reached its minimum when $y = -1$, i.e $g_{min} = g(-1) = -e^{-1}$, which is larger than $-0.5e$

So the inequality has no solution.

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  • $\begingroup$ Why not consider the (simpler) function $f(x)=e^x-2(x-1)$ (and use the same crucial step that $e^2\gt2$)? $\endgroup$ – Did Nov 12 '14 at 8:50
  • $\begingroup$ Thank you for your help! Appreciate it :) $\endgroup$ – jjepsuomi Nov 12 '14 at 8:51
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    $\begingroup$ @Did The inequality makes me think of Lambert W function. Of course there are others ways to do it $\endgroup$ – Petite Etincelle Nov 12 '14 at 8:52
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    $\begingroup$ @jjepsuomi You are welcome $\endgroup$ – Petite Etincelle Nov 12 '14 at 8:53

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