Learn How to Invert a Function with Mathematica for y = x^2

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Discussion Overview

The discussion revolves around inverting the function ##y = x^2## using Mathematica. Participants explore the implications of the function not being one-to-one and how that affects the existence of an inverse.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant suggests using the command Solve[y==x^2,x] in Mathematica to find the inverse.
  • Another participant points out that the function ##y = x^2## does not have an inverse that is itself a function because it is not one-to-one, as both ##x## and ##-x## yield the same output.
  • A later reply reiterates that Solve finds and reports both possible values for ##x##, which are ##\pm \sqrt{y}##.
  • One participant mentions checking with WolframAlpha, noting it is similar to Mathematica.

Areas of Agreement / Disagreement

Participants generally agree on the technical aspects of the function not being one-to-one and the implications for finding an inverse. However, there is a lack of consensus on how to interpret the results provided by Mathematica.

Contextual Notes

The discussion does not resolve the implications of using the inverse in practical applications or the limitations of the function's domain.

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?
 
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Solve[y==x^2,x]
 
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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?
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.
 
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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.
Solve finds and reports both.
 
Dale said:
Solve finds and reports both.
Right. I checked with wolframalpha, which is sort of the online version of Mathematica.
 

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