Intermediate Value Theorem

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Main Question or Discussion Point

Could someone please explain to me how I can use the intermediate value theorem to convince myself that m where

[tex] m = x^5 - 1 = x^3 [/tex]

exists. I have deduced that I can work in interval [1,2] using Newtons method where

[tex] x^5 - 1 - x^3 = 0 [/tex]

to find the answer. I even have the answer correct to four places, but part of the question is to show precisely how the intermediate value theorem proves the number exists in the positive real numbers before you do it.

My main problems are
1. I am not sure exactly how to prove that the function is continuous on the interval. (I have proved its continuous at each end.)
and
2. I am not sure exactly how to prove that the value for [itex] x^3 [/itex] is equal to the value for [itex] y^5 - 1 [/itex] when [itex] x=y [/itex] using **only** IVT. I mean, what in the intermediate value theorem proves that (and/or where) [itex] x^3 [/itex] crosses (or aligns with) [itex] x^5-1 [/itex]. Because I believe I am supposed to show that.

Just looking at the graphs of the equations and giving it a second thought is proving that the limit --> 0 exists on

[tex] x^5 - 1 - x^3 = 0 [/tex]

and is equal to f(0) hence proving continuity of the equation at x=0 what I should be doing?

using IVT then just on this equation would say that one value of x^5 - 1 is equal exactly to x^3. I think I have just answered my own question.
 
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Answers and Replies

  • #2
saltydog
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Well, a polynomial is continuous at every number: just take the limit of each monomial at the value a, and add them up to show that they equal f(a).

In regards to the Intermediate Value Theorem, it states that if the function is continuous on an interval [a,b], then it takes on all values between [f(a),f(b)]. Now, for some x<0, [itex]x^5-1[/itex] will be less than [itex]x^3[/itex]. Likewise, for some x>0, it will be larger. Thus, since they are continuous and as a consequence of the Intermediate Value Theorem, at some point they must cross, that is there exist an a such that:

[tex]a^5-1=a^3[/tex]
 
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  • #3
HallsofIvy
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In fact, since you yourself mention the interval [1, 2], what is f(x)= x5- x3- 1 when x= 1? What is f(2)? What does the intermediate value theorem tell you must happen between 1 and 2?
 
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saltydog said:
Well, a polynomial is continuous at every number: just take the limit of each monomial at the value a, and add them up to show that they equal f(a).
Thanks for that Saltydog. I could see easily that that was true, but I was not sure what rigorous proof was required to satisfy IVT.


saltydog said:
In regards to the Intermediate Value Theorem, it states that if the function is continuous on an interval [a,b], then it takes on all values between [f(a),f(b)]. Now, for some x<0, [itex]x^5-1[/itex] will be less than [itex]x^3[/itex]. Likewise, for some x>0, it will be larger. Thus, since they are continuous and as a consequence of the Intermediate Value Theorem, at some point they must cross, that is there exist an a such that:

[tex]a^5-1=a^3[/tex]

Thankyou again, I see that is one way that I can say it exists before I find it, but I am not sure I will naturally draw this conclusion in a test environment without fussing over details and flustering myself, since I am studying for a test I would probably find it easier to spell it out in the terms that Halls of Ivy recommended.
 
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