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Given $$g(x) = e^{-|x-x_0|/\alpha} + Ae^{-x/\alpha} + Be^{x/\alpha}$$ $\alpha >0$, $x_0 \in (0,1)$

Constraints are $$g(0) = g(1)$$ and $$g'(0) = g'(1)$$

Solve for $A$ and $B$.

My solution : $$A = \frac{e^{(x_0-1)/\alpha}}{1-e^{(-1/\alpha)}}$$ and $$B = \frac{e^{-x_0/\alpha}}{e^{1/\alpha}-1}$$

But when i verify, $g(0) \ne g(1)$. Certainly I am wrong and appreciate your help.

Edit : My differentiation $$g'(x) = -\frac{sign(x-x_0)}{\alpha}e^{-|x-x_0|/\alpha} -\frac{A}{\alpha}e^{-x/\alpha} + \frac{B}{\alpha}e^{x/\alpha}$$

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  • $\begingroup$ How do you differentiate $$g(x)$$ a function with absolute values? $\endgroup$ – Dr. Sonnhard Graubner Nov 21 '19 at 13:02
  • $\begingroup$ @Dr.SonnhardGraubner : $$g'(x) = -\frac{sign(x-x_0)}{\alpha}e^{-|x-x_0|/\alpha} -\frac{A}{\alpha}e^{-x/\alpha} + \frac{B}{\alpha}e^{x/\alpha}$$ $\endgroup$ – Rajesh Dachiraju Nov 21 '19 at 13:09

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