Answer:
Step-by-step explanation:
define the function:
[tex]g(x) = f(x) -x[/tex]
As both [tex]f(x)[/tex] and x are continuous functions, [tex]g(x)[/tex] will also be continuous.
Now, what can we say about [tex]g(a) = f(a) -a[/tex]?
we know that [tex]a\leq f(a) \leq b[/tex], thus:
[tex]a-a\leq f(a)-a \leq b-a\\0 \leq g(a) \leq b-a[/tex]
thus [tex] g(a) [/tex] is non-negative.
What about [tex]g(b)[/tex] ? Again we have:
[tex]a\leq f(b) \leq b\\a-b \leq f(b) -b \leq 0\\a-b \leq g(b) \leq 0[/tex]
That means that [tex]g(b)[/tex] is not positive.
Now, we can imagine two cases, either one of [tex]g(a)[/tex] or [tex]g(b)[/tex] is equal to zero, or none of them is. If either of them is equal to zero, we have found a fixed point! In fact, any point [tex]c[/tex] for which [tex]g(c)=0[/tex] is a fixed point, because:
[tex]g(c) = 0 \implies f(c) -c = 0 \implies f(c) = c[/tex]
Now, if [tex]g(a) \neq 0[/tex] and [tex]g(b) \neq 0[/tex], then we have that
[tex]g(a) >0[/tex] and [tex]g(b) < 0[/tex]. And by Bolzano's theorem we can assert that there must exist a point c between a and b for which [tex]g(c)=0[/tex]. And as we have shown before that point would be a fixed point. This completes the proof.
