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    $\begingroup$ You can extract the points from the plot and refine them using FindRoot. See, for instance, here. This is actually a pretty common method; there should be many examples of this on this very site. $\endgroup$ Commented Oct 3, 2016 at 18:13
  • $\begingroup$ @march Thank you, this helped me alot! I adopted the solution in this link to my case, with a few tweaks. Since I need a lot of precision, I'll post my tweaks below in case if others might be interested. $\endgroup$ Commented Oct 3, 2016 at 18:40
  • $\begingroup$ related . mathematica.stackexchange.com/q/75352/2079 $\endgroup$ Commented Oct 3, 2016 at 19:43
  • $\begingroup$ Just how do you want to use the "numerical intersection"? I don't see the purpose in having a few scattered points to very high precision. (1) For plotting, increasing precision beyond a few digits does not improve the image. (2) Given a point close the curve, should one keep x or y fixed in trying to improve its position, or should one move parallel to the gradient or some other direction? Your example fixes x, but that seems arbitrary and inappropriate for all possible curves. $\endgroup$ Commented Oct 4, 2016 at 15:24
  • $\begingroup$ @MichaelE2 my intent was to find the slope of the curve close to the spot where it bends back. Having a few numeric points in this subregion allows to do a fit. $\endgroup$ Commented Oct 6, 2016 at 14:24