An alpha particle in a nucleus is held by a potential having the shape shown in fig.Derive a function V(x) having this general form and having values V_0 and V_1 at x=0 and x=(+/-)x_1

Let us return to the problem.The fig. V(x) v/s x is given as an even function,symmetrical about y axis.At x=0, the curve intersects V(x) axis horizontally at V(x)=-V_0, where V_0>0. So at x=0, dV/dx=0.Then at |x|>0, the curve rises smoothly, crosses x axis, rises further, finally reaches a peak at x=x_1.The corresponding V(x)=V_1.[this part of the graph, in the (+ve) x side, is rather integration sign like, a bit tilted towards right,however].Then the curve falls like an exponential curve, towards infinity, never reaching zero.
So, you can see, there is a minimum at x=0 and two maxima at x=x_1
I tried hard to find V(x) and by trial and error, obtained the function as
V(x)=-V_0+x^2 [exp(-x^2)].This almost the same as above, as shown by my graphing software.But, it decays like rather damping vibration curve.as you can see, I do not need this.I have to have a function which cuts the x axis only twice, at x=(+/-)x_1.Please provide me with an appropriate function.

One more thing, how can I send picture in this forum?

Hello everyone.I have done it!
It is well verified after devised by me...
V(x)=(x^2-2)e^(-x^2)
however, I found that using (x^8-2) in the first place gives the curve an impression like the book.You may download a graphing software and try it out.