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

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The discussion centers on applying the Intermediate Value Theorem (IVT) to find a root of the function f(x) = x^3 + 2x + 1 within a specified interval of length 1/2. The initial attempt using the interval [0, 1/2] yielded values that do not bracket a root, prompting a reevaluation. A suggested interval of [-1/2, 0] shows promise, as f(-1/2) = -1/8 and f(0) = 1, indicating a sign change and thus the existence of a root by the IVT. The conversation highlights the importance of selecting intervals where the function values at the endpoints have opposite signs to confirm the presence of a root. Ultimately, the correct interval demonstrates the application of the IVT effectively.
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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".
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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