equation tangent line. $f_a(x)=(x-a)e^{a+2-x}.$

Question

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 :)

of course!!! I must find the point in which the tangent at this point meet $Oy$ in $A(0,2012)$ — Iuli, Dec 17 '13 at 01:00

score 0 · Answer 1 · answered Dec 17 '13 at 01:37

0

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?

answered Dec 17 '13 at 01:37

Mhenni Benghorbal

47,431

yes, I'm sure! :) – Iuli Dec 17 '13 at 07:46

equation tangent line. $f_a(x)=(x-a)e^{a+2-x}.$

1 Answers1