Don't undertstand what they are asking

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SUMMARY

The discussion focuses on applying the Intermediate Value Theorem to demonstrate the existence of a positive fifth root of 2. The function in question is defined as f(x) = x^5 - C, where C must be set to 2 to find the root. Participants clarify that the goal is to prove the existence of the root rather than simply identifying it, emphasizing the need to evaluate f(0) and f(2) to apply the theorem correctly.

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Homework Statement


The goal of this problem is to use the Intermediate Value Theorem to prove that there exists a positive number c which is the fifth root of 2.

Another way of expressing this is that we would like to find a positive root \,c of the continuous function f(x) = x^5 - ? .
Note: Fill in the box with an appropriate constant to complete the definition of the function f(x).


Homework Equations





The Attempt at a Solution


I tried 32, 1/32, which I thought they were just asking for 2^5? But I guess not.. I understand what the intermediate value theorem is, but not in the context of what they are asking.
 
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Chas3down said:

Homework Statement


The goal of this problem is to use the Intermediate Value Theorem to prove that there exists a positive number c which is the fifth root of 2.

Another way of expressing this is that we would like to find a positive root \,c of the continuous function f(x) = x^5 - ? .
Note: Fill in the box with an appropriate constant to complete the definition of the function f(x).


Homework Equations





The Attempt at a Solution


I tried 32, 1/32, which I thought they were just asking for 2^5? But I guess not.. I understand what the intermediate value theorem is, but not in the context of what they are asking.

You need to have f(x) = 0 when x = 2^{1/5}, not when x = 2^5.
 
The root of x^5-C is when x^5-C=0. You want to find some constant C such that the root is the fifth root of 2.
 
Oh got it, 2^1/5^5 = 2 so c = 2 thanks!
 
Chas3down said:
Oh got it, 2^1/5^5 = 2 so c = 2 thanks!

C=2, but I don't get your logic.
 
Chas3down said:
Oh got it, 2^1/5^5 = 2 so c = 2 thanks!

No, you are not done! You are assuming the existence of 2^(1/5), but you cannot do that: the question is asking you to prove that such a 5th root actually exists.
 
^ Listen to Ray. You haven't used the theorem! You may want to think about, say, f(0) and f(2), and how we can use these values and the theorem to show that 2 must have a positive fifth root.
 
johnqwertyful said:
C=2, but I don't get your logic.
He wrote "2^1/5^5" but meant (2^1/5)^5.
 

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