Use the intermediate value theorem to show that there exists a solution to x^5 + 2x 2 =0
I've read my notes on the IVT but still struggling. Could do with some help

Cggh90
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 03122010 17:20

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 03122010 17:46
You will have probably done something similar to this at Alevel. Let f(x)=x^5+2x2 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. 
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 03122010 18:11
(Original post by ttoby)
You will have probably done something similar to this at Alevel. Let f(x)=x^5+2x2 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.
So there is an x between 0 and 1? (0,1)
Is that it? 
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 03122010 18:31
(Original post by Cggh90)
Well if a=0, then f(a) < 0 and if b=1 then f(b)>0
So there is an x between 0 and 1? (0,1)
Is that it? 
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 03122010 21:14
did you understand the proof for it? I'm struggling.

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 03122010 21:49
(Original post by kfkle)
did you understand the proof for it? I'm struggling.
As a general sketch of the proof, if you have with a<b then you would look at the set
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.Last edited by ttoby; 03122010 at 21:50. 
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 03122010 22:19
(Original post by ttoby)
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/ma...sis2_notes.pdf 
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 04122010 00:16
(Original post by ttoby)
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/ma...sis2_notes.pdf
As a general sketch of the proof, if you have with a<b then you would look at the set
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. 
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 04122010 00:27
(Original post by kfkle)
I read the proof, thank you! just a small thing, does this proof show that c actually exists?
This part of the proof is in the two lines beginning "Note that S is...". 
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 04122010 00:35
(Original post by ttoby)
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 nonempty 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...".
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? 
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 04122010 02:56
(Original post by kfkle)
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?
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 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. 
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 04122010 03:08
(Original post by ttoby)
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 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.
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? 
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 04122010 15:36
(Original post by kfkle)
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?
I suppose different universities would teach things in a different order but hopefully you should cover sequences at some point. 
Cggh90
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 04122010 15:53
How about this question:
Use the IVT to show that there exists a solution to x^3 + ux  v =0, for all values of u,v > 0 ??
I know how to use the IVT normally now, but not for this one.
I need to find two bounds for x (a,b) where f(a) < 0 and f(b) > 0 but the u and v confuse me.. 
Cggh90
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 04122010 16:01
My attempt was this:
f(0) = v < 0
f(1) = 1 + u  v > 0 (as we are taking values when u = v ? ) 
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 04122010 17:12
(Original post by Cggh90)
My attempt was this:
f(0) = v < 0
f(1) = 1 + u  v > 0 (as we are taking values when u = v ? )
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+1v. 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. 
Cggh90
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 04122010 17:27
(Original post by ttoby)
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+1v. 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.
So f(v/u) = (v/u)^3 + v  v = (v/u)^3
(v/u)^3 is always greater than 0 as v and u are greater than 0 so interval (0,v/u)
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