maxima-bounces at math.utexas.edu wrote on 02/17/2011 11:45:56 AM:
> I defined a function giving a set of points with linearinterpol then I
added
> another function, which I defined with a simple formula. The resulting
> function is still strictly rising in value, so mathematically it can be
> inverted, so I wanted to invert the function, but I could not find an
> operator for that anywere, I tried
>
> simplify([y=f(x)],x), which produced some expression, but not what I
needed
Instead of simplify, maybe you need to use solve; try something like
(%i8) solve(y = 5 * x + 7,x);
(%o8) [x=(y-7)/5]
Thus the inverse of f(x) = 5 x + 7 is g(y) = (y-7)/5.
--Barton