Intermediate value theorem

You will have probably done something similar to this at A-level. Let f(x)=x^5+2x-2 so f is continuous.

Wat you need to do is find some a such that f(a)<0 and find some b such that f(b)>0. Then the intermediate value theorem will tell you that there is some x between a and b such that f(x)=0.

Well if a=0, then f(a) &lt; 0 and if b=1 then f(b)&gt;0

So there is an x between 0 and 1? (0,1)

Is that it?

did you understand the proof for it? I'm struggling.

It might help if you see it written in another way. I learnt it from the version on page 15 of http://www2.warwick.ac.uk/fac/sci/maths/people/staff/yossi_shamai/teaching/analysisii/analysis2_notes.pdf

It might help if you see it written in another way. I learnt it from the version on page 15 of http://www2.warwick.ac.uk/fac/sci/maths/people/staff/yossi_shamai/teaching/analysisii/analysis2_notes.pdf

As a general sketch of the proof, if you have $f(a)\leq0\leq f(b)$ with a<b then you would look at the set $S=\{x\in [a,b]:f(x)\leq 0\}$

So S is the set of all x values in [a,b] for which f(x) is less than or equal to zero. It might help if you sketch a graph of y=f(x), mark on the points a and b and then shade in the x values that are in S.

We show that x has a supremum c where c is the least upper bound of S

Intuitively, we would expect that if c is the sup of S then f(c) would be zero. But we need to prove this.

The proof is by contridiction. We show that if f(c)>0 then it doesn't work, if f(c)<0 then it doesn't work so we have to have f(c)=0.

I read the proof, thank you! just a small thing, does this proof show that c actually exists?

Yes, it uses the completeness axiom to do this (although the proof doesn't refer to that specifically, it is being used). It shows S is non-empty and S is bounded above by b. So hence (by completeness) S has a supremum, c.

This part of the proof is in the two lines beginning "Note that S is...".

That makes sense.

We were never actually given that particular axiom. Just to clear things up, upper bounds do exist in Q, but least upper bounds do not? (on open intervals)

and closed intervals do have least upper bounds in Q, correct?

I'm surprised you haven't covered completeness yet. You should have met it when you proved that Cauchy sequences are convergent.

There do exists open intervals on Q with a least upper bound, for example (0,1) has a least upper bound of 1. However the set $\{x\in\mathbb{Q}:x^2<2\}$ does not have a least upper bound. If it did then that least upper bound would be root 2 which isn't rational.

If you have a closed interval in Q of the form [a,b] then it would have a least upper bound: b as long as b is in Q.

hm yes i did realize the (0,1) example straight after I posted. thanks for the examples.

In any case, we haven't covered sequences at all! should we have? this is our first term, so it's pretty light on analysis. (the course title is calculus).

What year are you in?

My attempt was this:

f(0) = -v &lt; 0

f(1) = 1 + u - v &gt; 0 (as we are taking values when u = v ? )

But what if v=100 and u=1? Then f(1)<0.

However finding f(0) is a good start since we have a value of f that is guaranteed to be negative.

One approach I'm thinking of is to find f of something in terms of u and v so that the expression for f(x) is simpler. For example, f(1/u)=1/u^3+1-v. This eliminates the ux term in the middle but it isn't quite good enough because if v was very large then f(1/u) would be negative.

Try and find some value of x (in terms of u and v) so that the ux and the -v terms are both eliminated. Hint: try and find some value of x such that ux=v.

Then once you've found this value of x, it should be clear whether f(x) is positive or negative for this x.