What does it mean to say that a real valued function $ f : [a, b] \rightarrow \mathbb{R} $ is continuous at $ x_0 \in [a, b] $? Assume that $ f : [a, b] \rightarrow \mathbb{R} $ is continuous State, without proof, the Intermediate Value Theorem for $f$

Prove that there exist $x_1, x_2 ∈ [a, b]$ such that for all $ x \in [a, b]$ , $f(x_1)\leq f(x) \leq f(x_2)$.

[Standard results about sequences of real numbers may be used, without proof, provided they are clearly stated.]

This is a university question from a real analysis course. The way these questions go are they ask you to prove or state a particular theorem in the earlier part and then expect you to use that theorem to solve the problem in the next part. My initial thought to prove the last part was to state and prove the fact that a continuous function on a closed and bounded interval attains its bounds but that is not really in the spirit of the question which clearly wants use of IVT.

So,my attempt:

I split into cases:

1) $f$ was a constant function in which case choosing $x_1 = \dfrac{a+x}{2}$ and $x_2 = \dfrac{x+b}{2} $ would suffice.

2) $f(a)$ or $f(b)$ $\not=f(x)$ in which case you can use IVT on [a,x] or [x,b] one side for one of $x_1$ or $x_2$ and just let the other equal $x$ (I think this must be allowed to let $x=x_1$ $x=x_2$ sometimes otherwise you'd be in trouble with a function which had a maximum point or minimum in [a,b]- You wouldnt be able to find an $x_1$ or $x_2$ both distinct from $x$.I think the question probably means $x_1$ and $x_2$ are both distinct otherwise you could pick $x_1=x=x_2$ which would be stupid.

3) $f(a)$ and $f(b)$ both equal to $f(x)$ . $f$ is not constant so there is a value attained wich is greater than or less than $f(x)$. For the other side just let it equal $x$ again.

Is this fine? I have only used IVT for 2) and I have no idea what to do with the hint in the bracket about sequences of real numbers. If this is faulty or I have misunderstood something or there is an elegant way or any useful feedback I would appreciate it, thanks.


Sorry- I was right in first guess- it was just state and prove the fact that a continuous function on a closed and bounded interval attains its bounds. IVT has nothing to do with it. Use of IVT is for a further later part in the question which I have not included here.


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