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

In summary, by applying the Intermediate Value Theorem, a value c exists between -1/8 and 1 such that f(c) = 0 where f(x)=x3+2x+1.
  • #1
Dustobusto
32
0

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[itex]\in[/itex](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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  • #2
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[itex]\in[/itex](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.
 
  • #3
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:
  • #4
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".
 

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

1. What is the Intermediate Value Theorem?

The Intermediate Value Theorem is a mathematical principle that states that if a continuous function has two points with different values, then it must also have a point in between those two values. In other words, if a function starts at one value and ends at another, it must pass through all intermediate values.

2. How is the Intermediate Value Theorem used in mathematics?

The Intermediate Value Theorem is used to prove the existence of roots or solutions to equations. It is also used to prove the existence of extrema (maximum or minimum values) for a continuous function on an interval.

3. Can the Intermediate Value Theorem be applied to all functions?

No, the Intermediate Value Theorem only applies to continuous functions. A function is considered continuous if there are no sudden jumps or breaks in the graph and if it can be drawn without lifting the pencil from the paper.

4. What are some real-life applications of the Intermediate Value Theorem?

The Intermediate Value Theorem is used in various fields such as physics, engineering, and economics. For example, it can be used to determine the existence of a solution to a physical problem, such as the position of an object at a certain time, or the existence of a price for which a product will be sold.

5. Are there any limitations to the Intermediate Value Theorem?

Yes, there are some limitations to the Intermediate Value Theorem. It only guarantees the existence of a root or solution, but it does not provide a way to find it. Also, it only applies to continuous functions on a closed interval, so it cannot be used for all types of functions. Lastly, it cannot be used to find multiple roots or solutions within the same interval.

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