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How can I function range of the function x/(-1 + 2 x) over the domain (0 <= x<= 1)? I tried this but FunctionRange does not accept a domain like this.

FunctionRange[x/(-1 + 2 x), x, y, (0 <= x<= 1)]

EDIT: I wrongly wrote it as "function range of the function". Actually I means the range of the function.

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  • $\begingroup$ (0 <= x<= 1) is not a domain. From the FunctionRange docs, "Possible values for dom are Reals and Complexes. The default is Reals." $\endgroup$
    – Bob Hanlon
    Jan 19 at 23:56
  • $\begingroup$ You could use FindMaximum and FindMinimum.. $\endgroup$
    – bbgodfrey
    Jan 20 at 1:06
  • $\begingroup$ @bbgodfrey - to determine range using min and max you need to look at two intervals: Flatten[Outer[{#2, #1, #2[{x/(-1 + 2 x), #1}, x]} &, {0 <= x < 1/2, 1/2 < x <= 1}, {MaxValue, MinValue}], 1] // Grid[#, Frame -> All] & $\endgroup$
    – Bob Hanlon
    Jan 20 at 5:56
  • $\begingroup$ What do you mean by "How can I function range of the function..."? Is something missing? For example, "How can I find the function range of the function..."? Or something else? Preferably, please respond by editing (changing) your question, not here in comments (without "Edit:", "Update:", or similar - the question should appear as if it was written today). $\endgroup$ Jan 20 at 12:33

1 Answer 1

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Use {f, constraints} as the first argument of FunctionRange:

enter image description here

FunctionRange[{x/(-1 + 2 x), 0 <= x <= 1}, x, y]
y <= 0 || y >= 1
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