Finding the Optimal Value of Delta for Convergence in a Quadratic Function

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Homework Help Overview

The problem involves finding an optimal value for delta (d) in the context of a quadratic function, specifically f(x) = x² + x + 1, with given parameters a = 1 and L = 3. The goal is to establish a relationship between the distance from a and the function's output relative to L.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of the inequality 0 < |x - 1| < d and its relation to the function's output. There are attempts to manipulate the expression |f(x) - L| and to derive bounds for d. Some participants question the correctness of assumptions made in the derivation process.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's reasoning. Some guidance has been offered regarding the calculations, and there are multiple interpretations of the steps involved. No explicit consensus has been reached on the final value of d.

Contextual Notes

There are indications of potential errors in the calculations, particularly regarding the manipulation of inequalities and the resulting value of d. Participants are also navigating the constraints of the problem as they attempt to clarify their reasoning.

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Homework Statement



Suppose f(x) = x2 + x + 1, a = 1, and L = 3. Find a value d > 0 such that 0 < |x - a| < d implies |f(x) - L| < 1/100

Homework Equations


The Attempt at a Solution



Given 0<|x-1|<d implies 0<|x2 + x + 1 - 3|<1/100

0< x2 + x + -2 <1/100
0<(x+2)(x-1)<1/100

Assume
0<|x-1|<1
1<x<2
3<x+2<4

Then
3|x-1| < (x+2)|x-1| < 4|x-1| < 1/100

We need
4|x-1| < 1/100
|x-1| < 1/251/25 < 1, therefore d should be 1/25.
 
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zeion said:
Assume
0<|x-1|<1
1<x<2
3<x+2<4

In the second step it should be 0<x<2.
 
Ok.
But that doesn't really affect the rest does it?
 
The 3 should be a 2 in the next two statements following that. Also, d should be 1/400 because you divide 1/100 by 4. You accidently multiplied by 4. Other than that, it looks good.
 
Oh oops lol.
Ok thanks.
 

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