2. Show that the function, defined by f(x) : is NOT uniformly continuous on its x2 domain D(f) = {r € R: z>0}.
2. Show that the function, defined by f(x) : is NOT uniformly continuous on its x2 domain D(f) = {r € R: z>0}.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Using Example in photo1, solve the same way showing sidework(rightside) and proof(leftside) of the problem in photo2
Transcribed Image Text:### Problem 2:
Show that the function, defined by \( f(x) = \frac{1}{x^2} \) is **not uniformly continuous** on its domain \( D(f) = \{ x \in \mathbb{R} : x > 0 \} \).
### Explanation:
The task is to prove that the function \( f(x) = \frac{1}{x^2} \), where \( x \) is any positive real number, does not satisfy the condition of uniform continuity on its specified domain. Uniform continuity differs from ordinary continuity in that it requires the function to be uniformly continuous over its entire domain, not just at individual points.
![The image appears to contain handwritten mathematical notes, possibly related to an analysis or calculus concept about continuity.
---
**Transcription:**
1. \( f : [1, +\infty) \to \mathbb{R}, \quad f(x) = \frac{1}{x} \)
2. Then \( f \) is uniformly continuous on \( [1, +\infty) \).
---
**Proof:**
- Assume \( \varepsilon > 0 \).
- Let \( \delta \leq \varepsilon \).
- Consider \( x, u \in [1, +\infty) \).
- If \( |x - u| < \delta \), then:
\[
|f(x) - f(u)| = \left| \frac{1}{x} - \frac{1}{u} \right|
\]
- This can be simplified to:
\[
= \left| \frac{u - x}{x \cdot u} \right|
\]
- If \( x \geq 1 \) and \( u \geq 1 \), then it follows:
\[
\frac{1}{x \cdot u} \leq 1
\]
- Therefore:
\[
|x - u| < \delta \leq \varepsilon
\]
- Conclusively, \( f \) is uniformly continuous on \( [1, +\infty) \).
- \( \quad (\text{Q.E.D.}) \)
---
**Explanation:**
The notes detail a mathematical proof showing that the function \( f(x) = \frac{1}{x} \), defined on the interval \([1, +\infty)\), is uniformly continuous. The proof uses the definition of uniform continuity, primarily focusing on bounding differences using the given conditions and applying limits effectively. The calculus concepts of convergence, limits, and continuity seem central to the explanation provided.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F60239248-20bf-48ab-b0f3-ec79f0fa1bf1%2F8f8854f1-713b-4d6c-862a-fba5133a9163%2Fcdx3nub_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image appears to contain handwritten mathematical notes, possibly related to an analysis or calculus concept about continuity.
---
**Transcription:**
1. \( f : [1, +\infty) \to \mathbb{R}, \quad f(x) = \frac{1}{x} \)
2. Then \( f \) is uniformly continuous on \( [1, +\infty) \).
---
**Proof:**
- Assume \( \varepsilon > 0 \).
- Let \( \delta \leq \varepsilon \).
- Consider \( x, u \in [1, +\infty) \).
- If \( |x - u| < \delta \), then:
\[
|f(x) - f(u)| = \left| \frac{1}{x} - \frac{1}{u} \right|
\]
- This can be simplified to:
\[
= \left| \frac{u - x}{x \cdot u} \right|
\]
- If \( x \geq 1 \) and \( u \geq 1 \), then it follows:
\[
\frac{1}{x \cdot u} \leq 1
\]
- Therefore:
\[
|x - u| < \delta \leq \varepsilon
\]
- Conclusively, \( f \) is uniformly continuous on \( [1, +\infty) \).
- \( \quad (\text{Q.E.D.}) \)
---
**Explanation:**
The notes detail a mathematical proof showing that the function \( f(x) = \frac{1}{x} \), defined on the interval \([1, +\infty)\), is uniformly continuous. The proof uses the definition of uniform continuity, primarily focusing on bounding differences using the given conditions and applying limits effectively. The calculus concepts of convergence, limits, and continuity seem central to the explanation provided.
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