Question: If a function f(x) is defined for all real x and has an inverse f-1(x), does it necessarily follow that also g(x) = [f(x)2] has

If a function f(x) is defined for all real x and has an inverse f-1(x), does it necessarily follow that also g(x) = [f(x)2] has an inverse g-1(x)?
Explain why it does, or given an example when it does not.

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