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(limits) intermediate theorem 
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#1
Sep1408, 09:08 PM

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1. The problem statement, all variables and given/known data
use the intermediate theorem to show that the polynomial f(x) = x^3+3x2 has a zero in the interval [0,1]. 2. Relevant equations ???? 3. The attempt at a solution could someone explain to me where to start? when i graphed it, i saw that it intersects the x axis at about .6, but i still don't know how to find the answer using the theorem. i couldn't find any decent explanations in my book or on the internet. please help me to understand this theorem! 


#2
Sep1408, 09:51 PM

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What are f(0) and f(1)? What does the intermediate value theorem say?



#3
Sep1408, 10:17 PM

P: 23

f(0) = 2
f(1) = 2 so the zero must be located somewhere between 2 and 2? and since zero is located there, there is another value that can be used to pinpoint its location? at least that's how i comprehended the definition. 


#4
Sep1408, 10:21 PM

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(limits) intermediate theorem
You are getting the intermediate value theorem wrong. It doesn't say that the x such that f(x)=0 is between 2 and 2. It says that it is between 0 and 1. You don't need to pinpoint its location. Reread the theorem.



#5
Sep1408, 10:38 PM

P: 23

that's the problem, i can't figure out what it is saying...
"if f(x) is a continuous function on [a,b] then for every D between f(a) and f(b), there exists a C between a and b so that f(C) = D" i know that the function is continuous on [0,1], and for every value between f(0) and f(1), there is another value between 0 and 1 so that the function of the second value is equeal to the first value? i really am having trouble understanding the entire concept of what the definition implies. how can i use the theorem to actually show that the zero lies in [0,1]? could you perhaps give me an example or something? 


#6
Sep1408, 10:45 PM

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Ok, your example is the problem you've been given. You know it has a root near x=0.6, right? Take your statement of the intermediate value theorem and set a=0, b=1, f(a)=2 and f(b)=2. I think that turns into "if f(x)=x^3+3x2 is a continuous function on [0,1] then for every D between 2 and 2, there exists a C between 0 and 1 so that f(C) = D". Since 0 is between 2 and 2, there is a C so that f(C)=0. Do you see yet?



#7
Sep1408, 11:01 PM

P: 23

yeah, i think so... i was incorrect in using f(a) and a interchangably, as with f(b) and b.
it makes more sense the way you had it written. so my question is not actually asking me to find the exact coordinates where the zero is located, but rather to prove that the zero is located somewhere within the interval [0,1]. like you said, that can be proven because the theorem states that the zero is between 2 and 2 and thus there must be a point on the funtion that equeals 0 within the interval. i think i finally understand it reasonably well, but my head really hurts now :) 


#8
Sep1408, 11:06 PM

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S'ok. Give your head a rest. Forget the words. Take pencil and try to draw a continuous curve connecting (0,2) with (1,2) without crossing the x axis and x always increasing. Can't, right? That's all the theorem is saying.



#9
Sep1408, 11:10 PM

P: 23

thank's a lot! yeah, the words just made the math all the more confusing, but at least i get the concept now.
thanks for all of your help! 


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