kent davidge
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Suppose I have ##y = x^2##. By inverse, I mean ##x = \pm \sqrt y##. How can I get Mathematica to do that?
Solve[y==x^2,x]Technically speaking, ##y = x^2## doesn't have an inverse that is itself a function. For a function f to have an inverse, f must be one-to-one. Your example function fails this test because for each number x in the domain that maps to ##x^2##, -x also maps to the same value.kent davidge said:Suppose I have ##y = x^2##. By inverse, I mean ##x = \pm \sqrt y##. How can I get Mathematica to do that?
Solve finds and reports both.Mark44 said:Technically speaking, ##y = x^2## doesn't have an inverse that is itself a function. For a function f to have an inverse, f must be one-to-one. Your example function fails this test because for each number x in the domain that maps to ##x^2##, -x also maps to the same value.
Right. I checked with wolframalpha, which is sort of the online version of Mathematica.Dale said:Solve finds and reports both.