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.In your linear proof, specifically the a = 0 case, you used delta = epsilon, which worked just fine.  What are examples of two other choices of delta that would also have worked? 2.In your quadratic proof, is there any danger of this denominator being 0: |a|*(1 + 2|p|) + |b|?  Explain.

.In your linear proof, specifically the a = 0 case, you used delta = epsilon, which worked just fine.  What are examples of two other choices of delta that would also have worked? 2.In your quadratic proof, is there any danger of this denominator being 0: |a|*(1 + 2|p|) + |b|?  Explain.

Advanced Engineering Mathematics
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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My professor gave me these two questions about my proof. Please answer with plenty of details. Thanks!

1.In your linear proof, specifically the a = 0 case, you used delta = epsilon, which worked just fine.  What are examples of two other choices of delta that would also have worked?

2.In your quadratic proof, is there any danger of this denominator being 0: |a|*(1 + 2|p|) + |b|?  Explain.

 

= |(a(x + p) + b)(x - p)| = la(x – p) + blx – pl < la(x + p) + b|ô *< (lal - (2|pl + 1)+ |b|) - = E (lal - (2|p| + 1) + |b|) * la(x + p) + b| <|(x – p)| + lb| by Triangle Inequality = la|lx + p| + |b| < lal · (1x| + \pl) + |b| by Triangle Inequality = lal - (lpl +1+ |pl) + |b], Since |x| < \p] + ô = Ial • (2]p| + 1) + |비 > 이 Therefore, by the definition of continuity this shows g(x)is continuous at p e R.
Transcribed Image Text:= |(a(x + p) + b)(x - p)| = la(x – p) + blx – pl < la(x + p) + b|ô *< (lal - (2|pl + 1)+ |b|) - = E (lal - (2|p| + 1) + |b|) * la(x + p) + b| <|(x – p)| + lb| by Triangle Inequality = la|lx + p| + |b| < lal · (1x| + \pl) + |b| by Triangle Inequality = lal - (lpl +1+ |pl) + |b], Since |x| < \p] + ô = Ial • (2]p| + 1) + |비 > 이 Therefore, by the definition of continuity this shows g(x)is continuous at p e R.
Proof: Тopic: Linear Function: Determine whether a given function is continuous. We will prove that two functions, one linear and one Consider a linear function: f (x) = ax + b, such that a ER. quadratic, are continuous at specific points. We will show function f is continuous at c. Let's say e > Definitions: Сase 1: Definition of a limit: a + 0, Let's say ô = Ja Let IC R be an open interval, let c EI, let f:1- {c) - R be a function and let L So, we see if |x – cl < 6, then |f(x) - f(c) = lax + b - (ac + b)| = lax – ac| = |a(x – c) ER.The number L is the limit of f as x goes to c, written = lal · |x – c| < lal - 8 = |a · E. lim f(x) = L Therefore, by the definition of continuity this shows f is continuous at c. if for each e > 0, there is some 8 > 0 such that x EI- {c}and |x - c| < ô imply Case 2: If(x) – L| < E.If lim f(x) = L,we also say that f converges to L as x goes to c a = 0, Let's say 8 = ɛ If f converges to some real number as x goes to c, we say that lim f(x) exists. So, we see if |x – cl < 6, then |f (x) - f(c)| = |0x + b - (0x + b)| = 0 <E. Therefore, by the definition of continuity this shows f(x)is continuous at c E R. Definition of a continuous function: Let cE A.The function f is continuous at c if for each e > 0, there is some Quadratic Function: 8 > 0 such that x E A and |x – c| < 8 imply |f(x) – f(c)l < E. The function f Consider a quadratic function: g(x) = ax? + bx + c is discontinuous at c if f is not continuous at c; in that case we also say that such that a. b,CE Rare constants, a + 0, and p ER f has discontinuity at c. For ô > 0, x – pl < ô = -8 < x -p< 8 = -ô +p<x <p+ ô Let's say e > 0 Let's propose: Choose, 8 = min {1,Talalal 11u Ta (Note that the denominator is never zero.) A function f:1 → Ris continuous at cel if lim f(x) = f(c) "Ta|(2/pl +1) + 1B1} Now we see if |x – pl < 6, Then |g(x) – g(p)| = lax? + bx +c - (ap? + bp + c)| = la(x? – p2) + b(x – p)| = la(x +p)(x – p) + b(x – p)
Transcribed Image Text:Proof: Тopic: Linear Function: Determine whether a given function is continuous. We will prove that two functions, one linear and one Consider a linear function: f (x) = ax + b, such that a ER. quadratic, are continuous at specific points. We will show function f is continuous at c. Let's say e > Definitions: Сase 1: Definition of a limit: a + 0, Let's say ô = Ja Let IC R be an open interval, let c EI, let f:1- {c) - R be a function and let L So, we see if |x – cl < 6, then |f(x) - f(c) = lax + b - (ac + b)| = lax – ac| = |a(x – c) ER.The number L is the limit of f as x goes to c, written = lal · |x – c| < lal - 8 = |a · E. lim f(x) = L Therefore, by the definition of continuity this shows f is continuous at c. if for each e > 0, there is some 8 > 0 such that x EI- {c}and |x - c| < ô imply Case 2: If(x) – L| < E.If lim f(x) = L,we also say that f converges to L as x goes to c a = 0, Let's say 8 = ɛ If f converges to some real number as x goes to c, we say that lim f(x) exists. So, we see if |x – cl < 6, then |f (x) - f(c)| = |0x + b - (0x + b)| = 0 <E. Therefore, by the definition of continuity this shows f(x)is continuous at c E R. Definition of a continuous function: Let cE A.The function f is continuous at c if for each e > 0, there is some Quadratic Function: 8 > 0 such that x E A and |x – c| < 8 imply |f(x) – f(c)l < E. The function f Consider a quadratic function: g(x) = ax? + bx + c is discontinuous at c if f is not continuous at c; in that case we also say that such that a. b,CE Rare constants, a + 0, and p ER f has discontinuity at c. For ô > 0, x – pl < ô = -8 < x -p< 8 = -ô +p<x <p+ ô Let's say e > 0 Let's propose: Choose, 8 = min {1,Talalal 11u Ta (Note that the denominator is never zero.) A function f:1 → Ris continuous at cel if lim f(x) = f(c) "Ta|(2/pl +1) + 1B1} Now we see if |x – pl < 6, Then |g(x) – g(p)| = lax? + bx +c - (ap? + bp + c)| = la(x? – p2) + b(x – p)| = la(x +p)(x – p) + b(x – p)
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