considering x is an element of the real numbers and x^3 < 3
prove that there exists a real number 'alpha' such that
'alpha'^3 = 3
What does the first part ave to do with the proof?
Here's what I did:
By algebra of continuity, f(x) is continuous. You can prove this using the delta-epsilon definition.
By the intermediate value theorem s.t .
You can prove that is unique using the mean value theorem. It does not mention in the question that this is the case so I have no included it in my answer.
Hope this helps!
indeed, for this type of problem, what you did (using the intermediate value theorem) is more conventional. but it can also be done by the above method, in the first part of a beginning analysis class, before IVT even comes on the scene