Try this:
data=[1 2,
2 3,
3 3,
4 5,
7 9];
xval = @(yval) fsolve(@(x) interp1(data(:,1), data(:,2), x, 'linear') - yval, median(data(:,1)));
that with these calls to it:
Out3 = xval(3)
Out6 = xval(6)
Out8 = xval(8)
produces:
Out3 =
3.0000
Out6 =
4.7500
Out8 =
6.2500
