14. Suppose that a function f : R → R satisfies f(x+y) = f(x)+ f (y) for each x, y Є R. We already know that f(0) = 0, f is an odd function, and that f(nx) = nf(x) for all x = R and n ≥ Z. a) Prove that f(qx) = qf(x) for all x = R and q Є Q. b) Prove that f is continuous at 0 if and only if f is continuous on R. c) Prove that if f is continuous at 0, then there is an m Є R such that f(x) = mx for all x = R.
14. Suppose that a function f : R → R satisfies f(x+y) = f(x)+ f (y) for each x, y Є R. We already know that f(0) = 0, f is an odd function, and that f(nx) = nf(x) for all x = R and n ≥ Z. a) Prove that f(qx) = qf(x) for all x = R and q Є Q. b) Prove that f is continuous at 0 if and only if f is continuous on R. c) Prove that if f is continuous at 0, then there is an m Є R such that f(x) = mx for all x = R.
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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Question
Do part C only Don't use PMI only use given things.
![14. Suppose that a function f : R → R satisfies f(x+y) = f(x)+ f (y) for each x, y Є R. We already
know that f(0) = 0, f is an odd function, and that f(nx) = nf(x) for all x = R and n ≥ Z.
a) Prove that f(qx) = qf(x) for all x = R and q Є Q.
b) Prove that f is continuous at 0 if and only if f is continuous on R.
c) Prove that if f is continuous at 0, then there is an m Є R such that f(x) = mx for all x = R.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c836437-a2a4-49ad-8637-6abf33735fd4%2F77e71dbd-01a4-46e9-9106-020811e81c2d%2F6b6xzu_processed.png&w=3840&q=75)
Transcribed Image Text:14. Suppose that a function f : R → R satisfies f(x+y) = f(x)+ f (y) for each x, y Є R. We already
know that f(0) = 0, f is an odd function, and that f(nx) = nf(x) for all x = R and n ≥ Z.
a) Prove that f(qx) = qf(x) for all x = R and q Є Q.
b) Prove that f is continuous at 0 if and only if f is continuous on R.
c) Prove that if f is continuous at 0, then there is an m Є R such that f(x) = mx for all x = R.
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