Math Problem Statement

Below is the graph of the function f(x) = 1/x. find a positive number delta such that for all x satisfying 0 < | x - 1 | < delta, it holds that | f(x) - 1 | < 1/2. (if you prefer the string lingo, this question is essentially asking to find the width of the symmetric vertical strip around x = 1 such that for all x values from the vertical strip, the corresponding f(x) values lie in the horizontal strip shown in the graph. The width of this vertical strip is 2 times delta for some positive number delta, and you need to find delta.)

Solution

We are asked to find a positive number δ\delta such that for all xx satisfying 0<x1<δ0 < |x - 1| < \delta, it holds that f(x)1<12|f(x) - 1| < \frac{1}{2}, where the function is f(x)=1xf(x) = \frac{1}{x}.

Step-by-Step Solution:

We need to solve the inequality f(x)1<12|f(x) - 1| < \frac{1}{2}, or equivalently, 1x1<12\left| \frac{1}{x} - 1 \right| < \frac{1}{2}.

  1. Rewrite the absolute value inequality: 1x1=1xx<12\left| \frac{1}{x} - 1 \right| = \left| \frac{1 - x}{x} \right| < \frac{1}{2} This inequality simplifies to: 1xx<12\frac{|1 - x|}{|x|} < \frac{1}{2}

  2. Isolate 1x|1 - x|: Multiply both sides of the inequality by x|x| (which is positive for x>0x > 0): 1x<x2|1 - x| < \frac{|x|}{2}

  3. Consider the neighborhood around x=1x = 1: Since we are working around x=1x = 1, we expect xx to be close to 1. Let’s assume xx is close enough to 1 so that 12x32\frac{1}{2} \leq x \leq \frac{3}{2}. This is a reasonable assumption because for values of xx far from 1, the inequality won't hold. So within this range: |x| \approx 1 \quad \text{(since x is close to 1)}

    Using this approximation, the inequality simplifies to: 1x<12|1 - x| < \frac{1}{2}

  4. Find δ\delta: The condition 1x<12|1 - x| < \frac{1}{2} corresponds directly to δ=12\delta = \frac{1}{2}, since x1<δ|x - 1| < \delta describes the width of the vertical strip around x=1x = 1.

Thus, the value of δ\delta is 12\frac{1}{2}.

Conclusion:

A positive number δ\delta that satisfies the condition f(x)1<12| f(x) - 1 | < \frac{1}{2} for all xx such that 0<x1<δ0 < |x - 1| < \delta is δ=12\delta = \frac{1}{2}.


Do you have any questions or need further details?

Here are 5 related questions to consider:

  1. How would the value of δ\delta change if we required f(x)1<14|f(x) - 1| < \frac{1}{4}?
  2. What does the function f(x)=1xf(x) = \frac{1}{x} represent geometrically on a graph?
  3. Can we find the δ\delta for other values of f(x)f(x), such as f(x)=1x2f(x) = \frac{1}{x^2}?
  4. What would the behavior of f(x)f(x) be for large xx, and how does it affect δ\delta?
  5. How does the choice of δ\delta reflect the continuity of f(x)f(x) near x=1x = 1?

Tip: Whenever dealing with inequalities involving functions, rewriting the inequality in simpler terms can greatly help to visualize and solve the problem.

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Math Problem Analysis

Mathematical Concepts

Limits
Inequalities
Continuity
Reciprocal Function

Formulas

|1/x - 1| < 1/2
|(1 - x)/x| < 1/2
|1 - x| < |x|/2

Theorems

Epsilon-Delta Definition of Limits

Suitable Grade Level

Grades 11-12