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it might be a silly question but I tried everything and could not find the possible error.

I got

$$ f(x) = e^x $$

and I have to find all possible boundary points of $f(x)$ with tangent(s), which go through the point

$$ P (1/1) $$

Well, I'll just post what I did.

$$\begin{align} \frac{1-e^x}{1-x} &= f'(x) \\ \frac{1-e^x}{1-x} &= e^x \\ 1-e^x &= e^x - e^x\cdot x \\ 1+e^x\cdot x &= 2\cdot e^x \end{align}$$

...

edit: Thanks for the advice. Ok I'm stuck and I think on the wrong way.

Well I thought, the slope of that unknown tangent with $P(1/1)$ has to be the same as the derivative of the point I am looking for. $P$ obviously is not part of $f(x)$.

Maybe there's another way. I just need a hint. Thank you.

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    $\ln(1 + x\exp(x)) \ne \ln(1)+ \ln(x) + x$ Alpha finds only one solution to the equation above this, about $1.84141$ in terms of Lambert's W function.2011-12-13
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    im not sure what youre asking, but $\log(x+y)\neq\log x+\log y$ in general2011-12-13
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    @Ross, there must be two lines through $(1,1)$ tangent to the graph of $y=e^x$ (if that's what OP is asking).2011-12-13
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    @GerryMyerson: I see Bill Cook's result. When I typed it in, I only got the positive solution. Dunno why.2011-12-13

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