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Use the Intermediate Value Theorem to prove $f:[0,1]\to [0,1]$ continuous and $C\in[0,1]$, there is some $c \in [0,1]$ such that $f(c) = C$.

Using a similar technique to the proof of the intermediate value theorem, I can easily prove that there is an $f(x) = C$, but I am having trouble proving that a $f(c) = C$.

This is what I have:

Since $f$ is continuous on $[0,1]$ there exists $f(a) = 0$ and $f(b) = 1$.

Let $g(x) = F(x) - C$. We assume that $f(a) < C < f(b)$ $\to$ $0 < C < 1$

when $x = a$, $g(a)$ is negative when $x = b$, $g(b)$ is positive

Therefore, $g(a) < 0 < g(b)$, and since $f$ is continuous on $[0,1]$, so is $g$.

Therefore there exists a $g(x) = 0$, and $f(x) = C$.

How can I prove there is a $f(c) = C$ ? Is this a rule for IVT?

Thanks!

  • 1
    Ummm....wait, what? What is $c$ (as opposed to $C$)?2012-04-09
  • 3
    What you say that you can "easily prove", is wrong. Continuous does not imply that 0 and 1 are in the image.2012-04-09
  • 1
    In fact, the theorem as stated is not true. Consider $f(x) = x/2$ and $C=3/4$. You need tnat $f$ is onto $[0,1]$.2012-04-09

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