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The problem statement is as follows:

Let $y = f(x)$ be a continuous function defined on the closed interval $[0, b]$ with the property that $0 < f(x) < b$ for all $x \in [0, b]$. Show that there exists a point $c \in (0, b)$ with the property that $f(c) = c$. Hint: Consider the function $g(x) = f(x) - x$.

I have wrapping my head around this problem since a couple of days in my free time. Initially it seemed easy. Proving that for some $x \in (0, b)$, $f(x) = c$, is a no brainer. But actually proving that $f(c) = c$ has baffled me. So I thought pitching this question here.

Thanks in advance.

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Given $0 < f(x) < b $.

Notice that if $g(x) = x - f(x)$ then

$$ g(0) = - f(0) < 0 $$

while

$$ g(b) = b - f(b) > 0 \; \; \; \; ( since \; \; \; \; f(x) < b \; \; \forall x \in [0,b])$$

Now, use the IVT to conclude!

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    $\begingroup$ Thanks a lot. Attention to details. Kudos. $\endgroup$ – python_noob Oct 3 '18 at 18:39
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    $\begingroup$ But why did we choose this particular g(x) = f(x) - x? $\endgroup$ – python_noob Oct 3 '18 at 18:42
  • $\begingroup$ because it works! it does the trick! $\endgroup$ – Mikey Spivak Oct 3 '18 at 18:44

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