Invert function using InverseFunction and avoid discontinuity?

Questions
,
1

I want to invert this function:

f[x_] = FullSimplify[
  ConditionalExpression[
   x + (-70 + 21 x^2 + 
     Sqrt[7] Sqrt[
      700 + 3 x (-800 + 3 x (300 + x (-80 + 7 x)))])/(-140 + 42 x) - (
    x (-70 + 21 x^2 + 
       Sqrt[7] Sqrt[
        700 + 3 x (-800 + 3 x (300 + x (-80 + 7 x)))]))/(-140 + 42 x),
    0 <= x <= 1]]

I plot the function together with its inverse like this:

Plot[{f[x], InverseFunction[f][x]}, {x, 0, 1}, AspectRatio -> 1]

But this results in a discontinuous function. Is there a way to get around this using InverseFunction?
(I use Mathematica 11.0.1.0.)

example.gif
example.gif596×582 15.8 KB

2

InverseFunction must be used with circumspection, because it is not well defined, and it has no Reals option.
It seems that Reduce is wrong in your particular case too. Remove for simplicity the ConditionalExpression:

g[x_] = x + (-70 + 21 x^2 + 
      Sqrt[7] Sqrt[
        700 + 3 x (-800 + 3 x (300 + x (-80 + 7 x)))])/(-140 + 
      42 x) - (x (-70 + 21 x^2 + 
        Sqrt[7] Sqrt[
          700 + 3 x (-800 + 3 x (300 + x (-80 + 7 x)))]))/(-140 + 
      42 x);

Then the output of the following is wrong in Mma 11.0.1:

Reduce[y == g[x], x, Reals]

You can check it with a plot:

h[y_] = Piecewise[{{Root[-700 y + 
        700 y^2 + (1300 - 490 y - 420 y^2) #1 + (-2190 + 840 y + 
           63 y^2) #1^2 + (1920 - 399 y) #1^3 + (-804 + 63 y) #1^4 + 
        117 #1^5 &, 1], 
     y < 1}, {Root[-700 y + 
        700 y^2 + (1300 - 490 y - 420 y^2) #1 + (-2190 + 840 y + 
           63 y^2) #1^2 + (1920 - 399 y) #1^3 + (-804 + 63 y) #1^4 + 
        117 #1^5 &, 2], y >= 1}}];
ParametricPlot[{h[y], y}, {y, -1, 2}]
3

OK, thanks, helpful to know! I got the same result using Solve too.