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The following FunctionRange

(*a=((1-x) y)/(1-y z)*)
FunctionRange[{{((1 - x) y)/(1 - y z)}, 0 <= x <= 1, y > 1, 
   0 < z < 1}, {x}, a, Reals] // Simplify

returns

y > 1 && z < 1 && z > 0 && y z != 1 && a y z <= a && a <= y + a y z

I'm struggling to see how to illustrate/plot (in 2D or 3D) the values of x, y, z that are excluded and the resulting values for a that are excluded.

Appreciate any hints or tips.

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    $\begingroup$ Maybe result = FunctionRange[{((1 - x) y)/(1 - y z), 0 <= x <= 1, y > 1, 0 < z < 1}, {x}, a, Reals]; Reduce[result, a] $\endgroup$
    – cvgmt
    Aug 27, 2023 at 21:55
  • $\begingroup$ Apologies for the ambiguity. I've clarified the OP. What I meant by "illustrate the values" is some type of plot, 2D or 3D $\endgroup$
    – Hedgehog
    Aug 27, 2023 at 22:01

2 Answers 2

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a = ((1 - x) y)/(1 - y z);
RegionPlot3D[
 y > 1 && z < 1 && z > 0 && y z != 1 && a y z <= a && a <= y + a y z,
 {x, 0, 2}, {y, 0, 2}, {z, 0, 2},
 PlotPoints -> 100]

enter image description here

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  • $\begingroup$ Thanks, that led me to, what I believe is, the end result I was after. $\endgroup$
    – Hedgehog
    Aug 28, 2023 at 3:33
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If I understand correctly the following is the program:

  1. Calculate the FunctionRange in terms of the function a for one of the variables.
  2. Use Reduce to obtain a form suitable for plotting (HT: @cvgmt)
  3. Replace a to return to the original variables
  4. Again Reduce and then
  5. Plot the complement of the constraints using RegionPlot3D (HT: @david-p-stork)
  6. Some sort of color scale to show the values of a...
(*step 1*)
result = 
 FunctionRange[{((1 - x) y)/(1 - y z), 0 <= x <= 1, y > 1, 
   0 < z < 1}, {x}, a, Reals]
(*step 2, 3 and 4*)
r = Reduce[Reduce[result, a] /. a -> ((1 - x) y)/(1 - y z)]
(*step 5*)
RegionPlot3D[! (0 < z < 
     1 && ((1 < y < 1/z && 0 <= x <= 1) || (y > 1/z && 
        0 <= x <= 1))), {x, 0, 1}, {y, 1, 10}, {z, 0, 1}, ColorFunction -> "TemperatureMap",
 PlotPoints -> 50]

enter image description here

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