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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. –  Ross Millikan Dec 13 '11 at 0:17
    
im not sure what youre asking, but $\log(x+y)\neq\log x+\log y$ in general –  yoyo Dec 13 '11 at 0:17
1  
@Ross, there must be two lines through $(1,1)$ tangent to the graph of $y=e^x$ (if that's what OP is asking). –  Gerry Myerson Dec 13 '11 at 0:42
    
@GerryMyerson: I see Bill Cook's result. When I typed it in, I only got the positive solution. Dunno why. –  Ross Millikan Dec 13 '11 at 2:34
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1 Answer

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 $1.84141$.

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