Does the Intermediate Value Theorem Prove a Root in This Interval?

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Homework Help Overview

The discussion revolves around applying the Intermediate Value Theorem (IVT) to identify an interval of length 1/2 that contains a root of the function f(x) = x³ + 2x + 1. Participants are exploring the conditions under which the IVT can be applied, particularly focusing on the continuity of the function and the values at specific points.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss selecting intervals and evaluating the function at those points to determine if a root exists. There is a focus on ensuring that the function values at the endpoints of the interval have opposite signs, which is a requirement for the IVT to apply. Some participants express uncertainty about their choices and seek validation of their reasoning.

Discussion Status

There is ongoing exploration of different intervals, with some participants suggesting specific intervals like [-1/2, 0] and [-1/8, 1]. Guidance has been offered regarding the need for opposite signs at the endpoints to confirm the existence of a root, but no consensus has been reached on the best interval to use.

Contextual Notes

Participants are working under the constraints of needing to find an interval of a specific length (1/2) and are questioning their understanding of roots and the application of the IVT. There is mention of previous feedback from a tutor that has led to some confusion among participants.

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Homework Statement



Use the IVT to find an interval of length 1/2 containing a root of
f(x)=x3+ 2x + 1

Homework Equations



Intermediate Value Theorem: If f(x) is continuous on a closed interval [a, b] and f(a)≠f(b) then for every value M between f(a) and f(b) there exists at least one value c\in(a, b) such that f(c) = M

The Attempt at a Solution



So I am thinking with this what I need to do is take any 1/2 length interval and plug in those values for x. I took [0, 1/2] and plugged it in. I got

f(0) = 1

f(1/2) = 2.125 or 2 1/8. It just asked to find an interval. So I would think I could say f(c) exists somewhere between f(0) and f(1/2) because they are both continuous functions.

Let me know if this is right, because the math tutor told me it was wrong, and I think he's wrong.
 
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Dustobusto said:

Homework Statement



Use the IVT to find an interval of length 1/2 containing a root of
f(x)=x3+ 2x + 1

Homework Equations



Intermediate Value Theorem: If f(x) is continuous on a closed interval [a, b] and f(a)≠f(b) then for every value M between f(a) and f(b) there exists at least one value c\in(a, b) such that f(c) = M

The Attempt at a Solution



So I am thinking with this what I need to do is take any 1/2 length interval and plug in those values for x. I took [0, 1/2] and plugged it in. I got

f(0) = 1

f(1/2) = 2.125 or 2 1/8. It just asked to find an interval. So I would think I could say f(c) exists somewhere between f(0) and f(1/2) because they are both continuous functions.

Let me know if this is right, because the math tutor told me it was wrong, and I think he's wrong.

You are looking for a root, that is where f(c) = 0. So it would be good if your interval had f(x) having opposite signs at the ends.
 
root

I can't even remember roots. I guess I have to brush up on that.

By the way, just telling me that with that ONE SENTENCE was more than anything that particular tutor could do. I appreciate your succinctness.

Edit:

Ok, so to solve this, I could choose [-1/2, 0] ?

[-1/8, 1]

Zero exists between those two points, its a continuous function, so f(c)= 0 exists. Look alright?
 
Last edited:
Dustobusto said:
I can't even remember roots. I guess I have to brush up on that.

By the way, just telling me that with that ONE SENTENCE was more than anything that particular tutor could do. I appreciate your succinctness.

Edit:

Ok, so to solve this, I could choose [-1/2, 0] ?

[-1/8, 1]

Zero exists between those two points, its a continuous function, so f(c)= 0 exists. Look alright?

Yes, but you could phrase it better: "f(-1/2)=-1/8 and f(0) = 1 so by the intermediate value theorem there exists a c between -1/8 and 1 such that f(c) = 0".
 

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