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Let f: (0,1)→ R be a function and let a € (0,1). Match each statement in Group A with a statement from Group B which means the same thing. Group A: (i) Ve > 0,36 >0 such that |x-a| < 8 implies |ƒ(x) = f(a)| < €. (ii) €0, 0, x − a| < 8 implies [ƒ(x) − f(a)| < €. (iii) >0 such that V6 > 0, |x − a| < 8 implies |ƒ(x) − ƒ(a)| < €. (iv) 3€ >0 and 36 >0 such that |x − a| < 8 implies [ƒ(r) – f(a)| < €. (v) 0,3€ > such that |x − a| < 6 implies |ƒ(x) - f(a)| < €. (vi) 36 >0 such that Ve > 0, |x − a| < 6 implies |ƒ(x) - f(a)| < €. Group B: (a) f is continuous at a. (b) f is bounded on (0, 1). (c) f is constant on (0,1) (d) There is some neighbourhood of a on which f is bounded. (e) There is some neighbourhood of a on which f is constant.

Let f: (0,1)→ R be a function and let a € (0,1). Match each statement in Group A with a statement from Group B which means the same thing. Group A: (i) Ve > 0,36 >0 such that |x-a| < 8 implies |ƒ(x) = f(a)| < €. (ii) €0, 0, x − a| < 8 implies [ƒ(x) − f(a)| < €. (iii) >0 such that V6 > 0, |x − a| < 8 implies |ƒ(x) − ƒ(a)| < €. (iv) 3€ >0 and 36 >0 such that |x − a| < 8 implies [ƒ(r) – f(a)| < €. (v) 0,3€ > such that |x − a| < 6 implies |ƒ(x) - f(a)| < €. (vi) 36 >0 such that Ve > 0, |x − a| < 6 implies |ƒ(x) - f(a)| < €. Group B: (a) f is continuous at a. (b) f is bounded on (0, 1). (c) f is constant on (0,1) (d) There is some neighbourhood of a on which f is bounded. (e) There is some neighbourhood of a on which f is constant.

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter4: Writing Linear Equations
Section: Chapter Questions
Problem 7CA
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: 9. Let f (0,1)→ R be a function and let a € (0,1). Match each statement in Group A with a statement from Group B which means the same thing. Group A: (i) Ve > 0,38 >0 such that |x-a| < 8 implies |f(x) = f(a)| < €. (ii) Ve > 0, V6 > 0, x − a| < 8 implies |ƒf(x) - f(a)| < €. (iii) e > 0 such that V8 > 0, xa| < 6 implies f(x) = f(a)| < €. (iv) 3€ > 0 and 36 > 0 such that |x − a < 6 implies |f(x) - f(a)| < €. (v) V>0,3€ >0 such that |za| < 8 implies f(x) = f(a)| < €. (vi) 36 > 0 such that Ve > 0, x − a| < 8 implies [f(x) = f(a)| < €. Group B: (a) f is continuous at a. (b) f is bounded on (0, 1). (c) f is constant on (0,1) (d) There is some neighbourhood of a on which f is bounded. (e) There is some neighbourhood of a on which f is constant.
Transcribed Image Text:: 9. Let f (0,1)→ R be a function and let a € (0,1). Match each statement in Group A with a statement from Group B which means the same thing. Group A: (i) Ve > 0,38 >0 such that |x-a| < 8 implies |f(x) = f(a)| < €. (ii) Ve > 0, V6 > 0, x − a| < 8 implies |ƒf(x) - f(a)| < €. (iii) e > 0 such that V8 > 0, xa| < 6 implies f(x) = f(a)| < €. (iv) 3€ > 0 and 36 > 0 such that |x − a < 6 implies |f(x) - f(a)| < €. (v) V>0,3€ >0 such that |za| < 8 implies f(x) = f(a)| < €. (vi) 36 > 0 such that Ve > 0, x − a| < 8 implies [f(x) = f(a)| < €. Group B: (a) f is continuous at a. (b) f is bounded on (0, 1). (c) f is constant on (0,1) (d) There is some neighbourhood of a on which f is bounded. (e) There is some neighbourhood of a on which f is constant.
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