Math Problem Statement

Solve inequalities involving absolute values and find delta (δ) in terms of epsilon (ε).

Solution

You've provided a mathematical inequality and implication from a book:

x+4<ϵ212x+8<ϵ|x + 4| < \frac{\epsilon}{2} \Rightarrow |12x + 8| < \epsilon

Let's analyze and validate this implication:

  1. Understanding Absolute Values:

    • The absolute value x+4<ϵ2|x + 4| < \frac{\epsilon}{2} implies that the distance of x+4x + 4 from zero is less than ϵ2\frac{\epsilon}{2}. This can be rewritten as: ϵ2<x+4<ϵ2-\frac{\epsilon}{2} < x + 4 < \frac{\epsilon}{2}
    • By isolating xx, we get: ϵ24<x<ϵ24-\frac{\epsilon}{2} - 4 < x < \frac{\epsilon}{2} - 4
  2. Analyzing the Implication:

    • To see if this implies 12x+8<ϵ|12x + 8| < \epsilon, we substitute the bounds of xx into 12x+812x + 8: 12(ϵ24)+8and12(ϵ24)+812\left(-\frac{\epsilon}{2} - 4\right) + 8 \quad \text{and} \quad 12\left(\frac{\epsilon}{2} - 4\right) + 8
    • Simplifying these expressions: 12(ϵ24)+8=6ϵ48+8=6ϵ4012\left(-\frac{\epsilon}{2} - 4\right) + 8 = -6\epsilon - 48 + 8 = -6\epsilon - 40 12(ϵ24)+8=6ϵ48+8=6ϵ4012\left(\frac{\epsilon}{2} - 4\right) + 8 = 6\epsilon - 48 + 8 = 6\epsilon - 40
    • This results in: 6ϵ40<12x+8<6ϵ40-6\epsilon - 40 < 12x + 8 < 6\epsilon - 40
  3. Final Validation:

    • We need 12x+8<ϵ|12x + 8| < \epsilon, which translates to: ϵ<12x+8<ϵ-\epsilon < 12x + 8 < \epsilon
    • Given our bounds of 12x+812x + 8 as [6ϵ40,6ϵ40][-6\epsilon - 40, 6\epsilon - 40], the range does not immediately guarantee that it is within (ϵ,ϵ)(-\epsilon, \epsilon) for all values of ϵ\epsilon. The additional 40-40 term in the derived range can potentially cause it to exceed the desired bounds of ϵ-\epsilon to ϵ\epsilon, unless further conditions on ϵ\epsilon or xx are specified.

Therefore, the given implication x+4<ϵ212x+8<ϵ|x + 4| < \frac{\epsilon}{2} \Rightarrow |12x + 8| < \epsilon is not universally true without additional restrictions on ϵ\epsilon or xx.

Would you like more details on this topic or have any questions? Here are some related questions you might find useful:

  1. How do the terms in the inequality affect the range of solutions?
  2. What additional conditions on ϵ\epsilon or xx could make the implication always true?
  3. Can this type of implication be generalized for other coefficients or constants?
  4. How does scaling the variable xx in the absolute value affect the inequality's solution?
  5. What are the implications of altering the epsilon value in terms of real-world applications?

Tip: When working with inequalities and absolute values, always consider the effect of any constants or coefficients on the bounds of your solution.

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Math Problem Analysis

Mathematical Concepts

Absolute Value Inequalities
Epsilon-Delta Definition

Formulas

|x - a| < δ implies |f(x) - L| < ε
General form: |x - a| < δ, |f(x)| < ε

Theorems

Epsilon-Delta Limit Definition

Suitable Grade Level

Undergraduate Calculus