1 | initial version |

Another option is to use splines to approximate the curve. Here is some code inspired by https://stackoverflow.com/questions/31543775/how-to-perform-cubic-spline-interpolation-in-python

```
var('t y z')
P=desolve_system_rk4([z,-z^2*(-2+t/(1+t))-y],[y,z],ics=[0.1,10,2],ivar=t,end_points=100)
Q=[ [float(t1),float(z1)] for t1,y1,z1 in P]
x_points=([p[0] for p in Q])
y_points=([p[1] for p in Q])
tmp = interpolate.splrep(x_points, y_points)
f=lambda x: interpolate.splev(x, tmp)
```

You could then find a zero using:

```
find_root(f,12,20)
```

Also, you can compute the derivative using:

```
g=lambda x: interpolate.splev(x, tmp,der=1)
```

Then, find potential extrema using `find_root`

.

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