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Given a list of data, 

lst={{x1,y2},{x2,y2},...,{xn,yn}}

and the interpolation function fun,

y=Interpolation[lst]

which can be plotted as,

Plot[y[x],{x,x1,xn}]

as usual. I've tried to apply Solve in order to obtain x given y=y0 as

Solve[y[x]==y0,x]

and get

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >>

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  • $\begingroup$ Try InverseFunction[y]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented May 21, 2015 at 2:59
  • $\begingroup$ @Guesswhoitis. You mean InverseFunction[y][y0] , it does not work. I know it is a bit trivial, but maybe has to do with the fact it is an interpolation. $\endgroup$
    – Patrick El Pollo
    Commented May 21, 2015 at 3:05
  • $\begingroup$ That means your interpolating function is not one-to-one; that is only valid for one-to-one functions. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented May 21, 2015 at 3:29

2 Answers 2

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Method 1

You can reverse the value of {x,y} to {y,x}, and then interpolate them.

Note:In this case, the value of $y$ cannot be duplicate

lst=
 {{3.61648, 5.64818}, {7.53428, 4.52803}, {4.21088, 2.35117}, 
  {4.48224,1.08325}, {4.63735, 5.5877}, {2.24299, 3.10376}}

x = Interpolation[Reverse /@ data];
x[3.]

 2.44086

Method 2

If the the values of $y$ are be duplicate, you should interpolate in two directions.

For instance, using the following method

enter image description here

Options[interpolateCurve] = 
 Join[Options[ParametricPlot3D], Options[Interpolation]];

interpolateCurve[pts : {{_, _} ..}, opts : OptionsPattern[]] :=
  Module[{order, x, y, s, func1, func2},
   order = OptionValue[InterpolationOrder];
   x = pts[[All, 1]];
   y = pts[[All, 2]];
   (*calculate the accumulative chord length*)
   s =
    FoldList[
     Plus, 0, EuclideanDistance @@@ Partition[pts, 2, 1]];
   (*interpolation points with spline-method in two directions*)
   {func1, func2} =
    Interpolation[
     Thread@{s, #}, InterpolationOrder -> order, Method -> "Spline"] & /@ {x, y};
    (*visualize the curve*)
   ParametricPlot[{func1[t], func2[t]}, {t, 0, Last[s]},
    Evaluate@
     (Sequence @@ FilterRules[{opts}, Options[ParametricPlot]]), 
    Epilog -> {Red, PointSize[Medium], Point[pts]}]
]
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  • $\begingroup$ Method 1 worked. Thanks $\endgroup$
    – Patrick El Pollo
    Commented May 21, 2015 at 3:27
  • $\begingroup$ @Resanrom, My pleasure, and I will complete the method 2 when I have time. Now I am busy:-) $\endgroup$
    – xyz
    Commented May 21, 2015 at 3:28
  • 1
    $\begingroup$ @Resanrom, I have completed the method 2 $\endgroup$
    – xyz
    Commented Jul 13, 2015 at 8:36
  • $\begingroup$ .Thank you very much, your answers are really useful. I learn a lot. $\endgroup$
    – Patrick El Pollo
    Commented Jul 15, 2015 at 21:04
  • $\begingroup$ @xyz This is probably late, but for method 2, even though you can plot the inverse function, you can't actually extract the data points, that is, for a given y, you can obtain all the solutions x1,x2,x3,... such that y==foo[x]. Is there a work around? $\endgroup$
    – Andrew Yuan
    Commented Apr 15, 2020 at 4:43
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you can try FindRoot

f = Interpolation[{1, 2, 3, 5, 8, 5}];
sol = FindRoot[f[x] == 2.5, {x, 3}]
(*{x -> 2.5641}*)
Plot[{f[x], 2.5}, {x, 1, 6}, GridLines -> {{x /. sol}, None}]

enter image description here

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