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Taken from an old exam, I hope I don't translate it unclear.

Let $f: \left [ 0,1 \right ] \rightarrow \mathbb{R}$ continuous and let $f(\left [ 0,1 \right ]) \subset \left [ 0,1 \right ]$. By using intermediate value theorem, show that $f$ has at least got one fixed point in $\left[0,1\right]$ (aka there is one solution $x \in \left[0,1\right]$ of the equation $f(x) = x$).

I'm not sure at all if I did it right but I have taken the function:

$$f(x) = x$$

Then take the given intervals and insert for $x$ the beginning of interval:

$$f(0)=0$$

And the end of the interval:

$$f(1) = 1 >0$$

Because the function is continuous and because the equality sign changes to an inequality sign right after, the intermediate value theorem provides there must be at least one solution $x_{1} = [0,1]$.


Did I do it correctly? Did I explain correctly?

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  • $\begingroup$ How do you know $f(0) = 0$? Also, if you knew that then that would be your fixed point. $\endgroup$ – tilper Sep 21 '16 at 15:20
  • $\begingroup$ Because $f(x)=x$ put $x=0$ then it must equal $0$ $\endgroup$ – tenepolis Sep 21 '16 at 15:21
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    $\begingroup$ Right, I glossed over that part for some reason. Ok, but then why are you taking $f(x) = x$? Every point of that is a fixed point. $\endgroup$ – tilper Sep 21 '16 at 15:24
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    $\begingroup$ No. f(x) does NOT equal x. If that were so the question is trivial and all points are fixed points. x is NOT a variable in the statement "there is one solution x∈[0,1] of the equation f(x)=x)". In that statement $x$ is a value at which $f(x) = x$ but that is not the definition of $f$. And besides, it is what you need to prove; it's not a given. If it helps it might be better to read it as "Prove there is one solution $x_0 \in [0,1]$ where $f(x_0) = x_0$. $\endgroup$ – fleablood Sep 21 '16 at 15:27
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    $\begingroup$ I'm not going to "make answer" but I'll explain the question: $f$ is a continuous function. for all $x \in [0,1]$ we know $f(x) \in [0,1]$. That is ALL we know about $f$. Prove that there exists a $z \in [0,1]$ so that $f(z) = z$. $\endgroup$ – fleablood Sep 21 '16 at 15:30
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Hint: Apply the intermediate value theorem to the function $g(x)=f(x)-x$. Note that $g(0) \geq 0 \geq g(1)$.

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