Intermediate Value Theorem

[Kristina123]

Hi guys, I need some confirmation with my answers. I'm about 75% certain that my response is correct, but I'm having some doubts as the undefined value is throwing me off.

For 1a), f(2) = -5 and f(3) = 10. According to the IVT and assuming that this function is continuous, there must be a root x = c on [2,3] between 2 and 3, where f(c) = 0. Should I say that there "must"? There could be an undefined value between 2 and 3, for all we know right?

For 1b), f(3) = 10 and f(5) = -5. According to the IVT, we cannot guarantee that there is a root between 3 and 5 because at f(4) we have an undefined function value. We do not have enough information to prove that there is a root because the root may possibly be the point that is undefined at f(4).

Do you guys think my response is thorough and reasonably explained? Or am I missing some other details that could make this explanation a lot better?

Thank you!

 

[Otis]

Hi Kristina. I can understand your explanations well enough, but there are some things that could be reworded.

assuming that this function is continuous

We already know function f is not continuous (because it isn't defined at x=4). You ought to say, "assuming the function is continuous on [2,3]" because that's the interval you're working with in part (a).

a root x = c on [2,3] between 2 and 3, where f(c) = 0.

It's okay to call c a root, but then it's redundant to add that f(c) is zero. You could replace the word 'root' with 'value'. That way, your statement f(c)=0 serves to identify value c as the root.

Another redundancy is to say "on [2,3] between 2 and 3" because those are both the same interval (endpoints notwithstanding). Delete the "on [2,3]" part and simply say "between x=2 and x=3".

Should I say that there "must"? There could be an undefined value between 2 and 3, for all we know right?

There are no undefined values of f(x) between x=2 and x=3 because we have already assumed the function is continuous on that interval.

And there's nothing wrong with using the word 'must'. As x goes from 2 to 3, we know that f(x) -- that is, y -- takes on every value ranging from -5 to 10 because f is continuous there. Zero is one of those y-values. So, yes, f(x) must be 0 at some x-value within the interval [2,3].

at f(4) we have an undefined function value. We do not have enough information to prove that there is a root because the root may possibly be the point that is undefined at f(4).

Again, your intent is clear, but we can tweak the wording a bit. The symbol f(4) doesn't represent anything in this exercise, so it's better to not use it. Instead, you may say, "at x=4 the function is not defined".

Lastly, we don't want to suggest that a root could be at a point that's undefined. A better explanation would be that a root is not guaranteed because the "discontinuity might be at (4,0)".

Overall, your answers are a fine first draft. And it's a good sign that you'd recognized a need to inquire about the wordings, too.

Keep up the good work.

?

 

[Steven G]

No, no no!
I do not agree with your answers at all. All you know is that f(2) = -5, f(3) = 10, f(5) = -5 and f(4) is undefined. Did you graph these 3 points ((2,-5), (3, 10), (5,-5) )?? It is very very possible that f(x) ONLY contains these 3 points and nothing else. It is possible that f(x) has a few more points or maybe it is just discontinuous at x=4.
You can ONLY use the Intermediate Value Theorem on intervals where the function is known to be continuous. You have NO IDEA AT ALL if f(x) is continuous anywhere!

The answer to part a and b is Not Possible.

@Otis, please comment on my post.

 

[Otis]

maybe [f(x)] is just discontinuous at x=4

Yes, that's what I've assumed (partly because we're on the calculus board), and perhaps I ought to have advised the OP to also state an assumption of continuity in the other interval, too, except at x=4.

You have NO IDEA AT ALL if f(x) is continuous anywhere!

please comment

With pleasure, Jomo. They say the table provides information about function f. I'd expect them to say something else, if the table's meant to define f(x) as a function of discreet values.

Plus, if a function domain contains only a few numbers, then I don't see a need to pick one number outside the domain just for specifying in a list of three valid (x,y) pairs that the function is not defined at that one number. An assumption that f is continuous except at x=4 is what makes sense to me, and assumptions would be included in my answers.

I do agree with your answers, assuming a function domain of discreet values, of which only three are listed. IVT would not apply.

?