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Calculus Analysis proof

  1. Jul 16, 2010 #1


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    1. The problem statement, all variables and given/known data
    [tex]g[/tex] is a continuous and differentiable function on[tex](a,b)[/tex]. Prove that if for all [tex]x\in(a,b)[/tex], [tex]g'(x)\neq0\Rightarrow[/tex] g is an injective function.

    Is there anything wrong about my proof? I'm not comfortable in making logical mathematical arguments: I usually make small inaccuracies. Or do a lot of work going down a bad path Any help appreciated.

    3. The attempt at a solution

    We select an arbitrary sub-interval on [tex](a,b)[/tex]. We call it [tex](m,n)[/tex] We know that g is also continuous and differentiable on any sub-interval of the interval that it is continuous and differentiable on.

    The "injective functoin" has the definition that [tex]f(a)=f(b)\Rightarrow a=b[/tex] That is, no two values on the domain can map to the same point on the co-domain.

    So for now we shall assume that[tex]f(m)=f(n)[/tex]. This allows us to make use of the Mean value theorem which states that if a function is continuous and, differentiable on an interval, in this case [tex](m,n)[/tex] and [tex]f(m)=f(n)[/tex] then [tex]\exists c\in (m,n)[/tex] such that [tex]g'(c) = 0[/tex]. This leads to a contradiction as [tex]g'(x)\neq0[/tex]. Therefor no such interval [tex](m,n)[/tex] can exist. Hence [tex]f(m)=f(n)\Rightarrow m=n[/tex] (The point m on the domain is equivalent to(the same point as) n) and [tex]g[/tex] is injective.
  2. jcsd
  3. Jul 16, 2010 #2


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    The proof is ok. Your phrasing at the beginning is pretty awkward. You aren't selecting (m,n) to be an arbitrary sub-interval. You are doing a proof by contradiction. So you assume f is NOT injective. If f is NOT injective then there are two different points m and n such that f(m)=f(n). Now the rest of the proof is fine.
  4. Jul 17, 2010 #3


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    Thanks very much!
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