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I've managed to find the derivative to be $f'(x) = \frac{1+e^{x-3} - x(e^{x-3})}{(1+e^{x-3})^2} = 0$

I'm posting a new example because I've noticed too many mistakes in the way I wrote the question that it's easier to scratch it.

I'm stuck on this part but I'm also unsure whether it's correct to do so:

$1+e^{x-3} - x (e^{x-3}) = 0$

Seth
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nyugnep
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  • It's not $x=0$, Using MatLab I approximated it to be either $x = 2.5571$ or $x = -2.5571$. I just need to find the exact value using pen and paper – nyugnep Feb 28 '19 at 01:05
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    This equation can not be solved algebraically. You might use a numerical approach, like the Newton's method. – Cornman Feb 28 '19 at 01:11
  • @nyugnep You have to solve $e^{x-3}=\frac{1}{x-1}.$ Plot both curves. There is only one solution, which is positive. For an estimation you need a numerical approach as claims Cornman5. – user376343 Feb 28 '19 at 06:50

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