Does the function have a zero in the given interval?

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The discussion centers on verifying the existence of a zero for the function f(x) = x³ - 4x + 2 within the interval (1, 2) using the Intermediate Value Theorem (IVT). Participants confirm that f(1) = -1 and f(2) = 2, establishing that a zero exists due to the change in sign. The method of subdividing the interval into 10 subintervals is recommended to approximate the zero to three decimal places, with a focus on applying the Bisection Method for further precision.

PREREQUISITES
  • Understanding of the Intermediate Value Theorem
  • Familiarity with polynomial functions
  • Knowledge of the Bisection Method for root approximation
  • Basic skills in evaluating functions at specific points
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  • Study the application of the Intermediate Value Theorem in various contexts
  • Learn the Bisection Method in detail for root-finding
  • Practice evaluating polynomial functions at multiple points
  • Explore numerical methods for approximating roots of equations
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Students in calculus, mathematicians interested in numerical methods, and educators teaching root-finding techniques will benefit from this discussion.

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.
 
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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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