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Let $f$ be the following function $$f_a(x)=(x-a)e^{a+2-x}.$$ I have to determine a point where the tangent in this point meet $Oy$ axis in point $A(0,2012)$.

I made $f'(x)=(1-x+a)e^{a+2-x}$. and then equation of tangent:

$y=f(b)+f'(b)(x-b)$ $$y=(b-a)e^{a+2-b}+(1-b+a)e^{a+2-b}(x-b)$$. now I make $y=2012$ and $x=0$ but i don't know how to find $b$.

thanks :)

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  • $\begingroup$ Does the point lie on the curve? $\endgroup$
    – Mhenni Benghorbal
    Dec 17, 2013 at 0:51
  • $\begingroup$ Can you post the original question? $\endgroup$
    – Mhenni Benghorbal
    Dec 17, 2013 at 0:57
  • $\begingroup$ of course!!! I must find the point in which the tangent at this point meet $Oy$ in $A(0,2012)$ $\endgroup$
    – Iuli
    Dec 17, 2013 at 1:00
  • $\begingroup$ @luli: Ok. That was my understanding. $\endgroup$
    – Mhenni Benghorbal
    Dec 17, 2013 at 1:03

1 Answer 1

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Assume the point is $(x_0,y_0)$, then it satisfies the curve equation

$$ y_0 = (x_0-a)e^{2-a+x_0} $$

and tangent equation

$$ y_{{0}}-2012=-{{\rm e}^{a+2-x_{{0}}}} \left( -1+x_{{0}}-a \right) x_{{0 }} . $$

Now, you need to solve the two equations. By the way, it is not easy to solve this system. So, are you sure of your function?

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  • $\begingroup$ yes, I'm sure! :) $\endgroup$
    – Iuli
    Dec 17, 2013 at 7:46

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