Does the function have a zero in the given interval?

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Discussion Overview

The discussion revolves around verifying whether the function f(x) = x³ - 4x + 2 has a zero within the interval (1, 2) and applying the Intermediate Value Theorem to approximate this zero. Participants explore the method of subdividing the interval into smaller segments to find the zero, while also expressing confusion about the instructions and the theorem itself.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation

Main Points Raised

  • Some participants express confusion regarding the instructions for using the Intermediate Value Theorem and how to subdivide the interval correctly.
  • One participant calculates f(1) and f(2) to demonstrate that the function changes sign over the interval, suggesting that a zero exists according to the Intermediate Value Theorem.
  • Another participant proposes that dividing the interval into subintervals might imply using the Bisection Method to refine the approximation of the zero.
  • Several participants mention their experiences with the Bisection Method, indicating it can be time-consuming.
  • There are references to previous discussions and posts on other math forums, indicating a broader conversation on the topic.

Areas of Agreement / Disagreement

Participants generally agree that the function has a zero in the interval based on the Intermediate Value Theorem, but there is no consensus on the best method to approximate the zero or on the interpretation of the instructions regarding subdivision.

Contextual Notes

Some participants note the potential confusion in the instructions and the varying interpretations of how to apply the Intermediate Value Theorem and the subdivision process.

Who May Find This Useful

This discussion may be useful for students learning about the Intermediate Value Theorem, the Bisection Method, and those seeking clarification on how to approach problems involving function zeros within specified intervals.

nycmathdad
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Verify the given function has a zero in the indicated interval. Then use the Intermediate Value Theorem to approximate the zero correct to three decimal places by repeatedly subdividing the interval containing the zero into 10 subintervals.

f (x) = x3 − 4x + 2; interval: (1, 2)

I don't understand the instructions. How is this done?
 
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Beer soaked ramblings follow.
nycmathdad said:
Verify the given function has a zero in the indicated interval. Then use the Intermediate Value Theorem to approximate the zero correct to three decimal places by repeatedly subdividing the interval containing the zero into 10 subintervals.

f (x) = x3 − 4x + 2; interval: (1, 2)

I don't understand the instructions. How is this done?
Repost from https://mathforums.com/threads/verify-given-function.356475/#post-644773 by same guy as nycmath, https://mathhelpboards.com/members/rtcntc.8471/, https://mathhelpboards.com/members/harpazo.8631/, https://mathhelpboards.com/members/mathland.19617/.
Read the relevant section of your book.
 
nycmathdad said:
Verify the given function has a zero in the indicated interval. Then use the Intermediate Value Theorem to approximate the zero correct to three decimal places by repeatedly subdividing the interval containing the zero into 10 subintervals.

f (x) = x^3 − 4x + 2; interval: (1, 2)

I don't understand the instructions. How is this done?
Do you know what the "Intermediate Value Theorem" is?

f(1)= 1- 4+ 3= -1 and f(2)= 8- 8+ 2= 2. The Intermediate Value Theorem says that since this polynomial is continuous and positive at one end of the interval and negative at the other end there must be some point in the interval where f(x)= 0.

Now, divide (1, 2) into 10 subintervals:
(1, 1.1), (1.1, 1.2), (1.2, 1.3), (1.3, 1.4), (1.4, 1.5), (1.5, 1.6), (1.6, 1.7), (1.7, 1.8), (1.8, 1.9), and (1.9, 2.0).
Evaluate f(x) at the endpoints of those. If there is an interval where f(x) has different signs at the endpoints then, by the "Intermediate Value Theorem", f(x)= 0 somewhere in that interval. Divide that interval into 10 more subintervals and repeat. That will get you to things like 1.a0, 1.a1, 1,a2, etc, two decimal places. Repeat one more time to get three decimal places.
 

I have a feeling that when it says to divide into subintervals, they are meaning to Bisect it ten times to close in on the point...
 
Prove It said:
I have a feeling that when it says to divide into subintervals, they are meaning to Bisect it ten times to close in on the point...

Yes but I am beyond this topic in my studies.
 
nycmathdad said:
Yes but I am beyond this topic in my studies.

Are you saying you've already done the Bisection Method? Good, then attempt it.
 
Prove It said:
Are you saying you've already done the Bisection Method? Good, then attempt it.

I have done several questions involving the Bisection Method. It is very time consuming.
 
nycmathdad said:
I have done several questions involving the Bisection Method. It is very time consuming.
Not if you have someone else do it for you!
 
Country Boy said:
Not if you have someone else do it for you!

I did a few myself. Struggled but actually did it.
 
  • #10
Beer soaked ramblings follow.
nycmathdad said:
I did a few myself. Struggled but actually did it.
Translation: I posted actual examples on other math sites and inveigled others to do it for me.
 

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