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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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    @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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The equation of the tangent to $y=e^x$ at the point $x=a$ is $y-e^a = e^a(x-a)$ since the tangent has slope f'(a)=e^a and passes through the point $(a,e^a)$.

I surmise your question is "When do the tangents to $y=e^x$ pass through the point $(1,1)$?"

This occurs exactly when $(x,y)=(1,1)$ satisfies the equation $y-e^a=e^a(x-a)$. Thus we must have $1-e^a=e^a(1-a)$. Thus $1-e^a=e^a-ae^a$ and so $2e^a-ae^a-1=0$. This is a non-linear equation whose solution is given by Wolfram Alpha to be $-1.14619$ and $1.84141$.