Thread: Trouble solving a question: lim[x->-infty]f(x) * lim[x->+infty] < 0; find max of g(x)

1. Originally Posted by Dr.Peterson
"since I didn't use the theorem explicitly, only its idea, it seemed ok at the time": If you were told to prove your result, then you have to give clear reasons (though the level of detail depends on the course). I imagine the real issue may be that you didn't use the theorem explicitly. If you were just told to solve the problem, on the other hand, then it wouldn't really be necessary.

"the teachers' reference is only that I didn't prove the interval is bounded": that still seems odd if you didn't even mention an interval, as you indicated. But there may be a language issue here.

"continuous in a closed [a,b] (doesn't mention bounded)": An interval [a,b] is by definition bounded; they don't have to say that separately. I'm almost sure what your teacher said was that you didn't prove that it is continuous on a closed, bounded interval, which amounts to saying "on some [a,b]".

But you'll have to ask your teacher for a fuller explanation, since we can't know just what was said, much less what was meant. Teachers want to be understood, so asking is always a good thing.

If you do find out what is going on, some of us would probably like to hear about it ...
Thank you for your help. The teacher's issue is that you absolutely had to prove such an interval was also bounded which, to me, is odd, since it's closed I assume it has to be bounded.

There have been a number of issues with this teacher throughout the course, so I just wanted to confirm that the interval was already bounded. Now I know what to do. Thanks again!

I'll come back when I have further developments on the issue!

2. Not sure I understand all the closed vs. bounded business, but if you need to prove that [a,b] exists such that f(a) > 0 and f(b) < 0 I would try something along these lines:

Assume if x-> +inf f(x) -> A, A > 0. Let's assume there is NO such a, that f(a) > 0. By definition of lim f(x) for any epsilon we should be able to get f(x) closer to A than epsilon: |f(x) - A| < eps.
Let's make A the epsilon:
|f(x) - A| < A
f(x) - A < A if f(x) > A - contradicts our assumption that at no a f(a) > 0.
A - f(x) < A if f(x) < A =>
=> -f(x) < 0 => f(x) > 0 - again contradicts our assumption that at no a f(a) > 0.
Therefore, there should be an a where f(a) > 0.

Same for b and the case when limits are reversed.
Sorry if this is sloppy, took calculus 20+ years ago

3. Originally Posted by lev888
Not sure I understand all the closed vs. bounded business, but if you need to prove that [a,b] exists such that f(a) > 0 and f(b) < 0 I would try something along these lines:

Assume if x-> +inf f(x) -> A, A > 0. Let's assume there is NO such a, that f(a) > 0. By definition of lim f(x) for any epsilon we should be able to get f(x) closer to A than epsilon: |f(x) - A| < eps.
Let's make A the epsilon:
|f(x) - A| < A
f(x) - A < A if f(x) > A - contradicts our assumption that at no a f(a) > 0.
A - f(x) < A if f(x) < A =>
=> -f(x) < 0 => f(x) > 0 - again contradicts our assumption that at no a f(a) > 0.
Therefore, there should be an a where f(a) > 0.

Same for b and the case when limits are reversed.
Sorry if this is sloppy, took calculus 20+ years ago
Yes, that's just about exactly what I was hoping the OP would either say, or realize the need for. But without knowing what the teacher actually said, we can only suppose it's what the teacher thinks is missing. I think it is. And perhaps I should have given a more direct hint in this direction.

Your work could also be expressed without proof by contradiction, I believe. But your thinking is good.

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