I'm trying to define a function f(x) such that 1/f(x) = 1/R - A/B. I know that I can just work the RHS out so that f(x) = C/D and then define it like that, but I was wondering if there was a way to directly define the function without simplification. In some cases it may be very difficult to simplify it to get a f(x) = form.

  • 2
    $\begingroup$ Your code-formatted bits are not Mathematica code -- just to confirm, you want to do this is in the software Mathematica? $\endgroup$
    – Michael E2
    Apr 22, 2021 at 0:59
  • $\begingroup$ f[x_]:=1/(1/R - A/B) ? $\endgroup$
    – murray
    Apr 22, 2021 at 14:27
  • $\begingroup$ @MichaelE2 yes its in Mathematica $\endgroup$
    – sarat.kant
    Apr 22, 2021 at 15:42
  • $\begingroup$ @murray yes like I mentioned, this might not be simple to do in other situations so I wanted a way to define it directly $\endgroup$
    – sarat.kant
    Apr 22, 2021 at 15:42
  • $\begingroup$ Are you looking for a function inverse or a reciprocal? $\endgroup$ Apr 23, 2021 at 0:18

1 Answer 1


You can use InverseFunction for this so long as your LHS is of the form g[f[x]]:

g[y_] := 1/y;
f[x_] := InverseFunction[g] @ (1/R[x] - A[x]/B[x])

If InverseFunction[g] can be computed symbolically, then it might be more efficient to use immediate assignment for f instead (i.e, f[x_] = ) so that the inverse does not have to be re-computed each time you call f, but you'll need to be careful that x is not already defined then. You can use Block or Module to ensure this, if necessary:

 f[x_] = InverseFunction[g] @ (1/R[x] - A[x]/B[x])

If you don't like defining an axillary function g, you can use pure functions instead (e.g., InverseFunction[1/# &]).


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