Consider the real-valued function of one variable f: R → R defined by - {₁+2²+ (a) Show that f'(0) f(t)= = for t= 0 +2t² sin () for t = 0. = 1, and that the derivative f'(t) exists and is bounded in the open interval (-1,1). (b) Show that f'(t) is not continuous at t = 0. (c) Show that f is not one-to-one in any neighborhood of 0 (and hence not invertible there). [This shows that, even in the one-dimensional case n = 1, the continuity of the deriva- tive f' at the point a is needed in the Inverse Function Theorem, even if f'(a) #0!]
Consider the real-valued function of one variable f: R → R defined by - {₁+2²+ (a) Show that f'(0) f(t)= = for t= 0 +2t² sin () for t = 0. = 1, and that the derivative f'(t) exists and is bounded in the open interval (-1,1). (b) Show that f'(t) is not continuous at t = 0. (c) Show that f is not one-to-one in any neighborhood of 0 (and hence not invertible there). [This shows that, even in the one-dimensional case n = 1, the continuity of the deriva- tive f' at the point a is needed in the Inverse Function Theorem, even if f'(a) #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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![Problem 1
Consider the real-valued function of one variable f: R → R defined by
So
f(t)= =
for t= 0
t+2t² sin () for t = 0.
(a) Show that f'(0) = 1, and that the derivative f'(t) exists and is bounded in the
open interval (-1,1).
(b) Show that f'(t) is not continuous at t = 0.
(c) Show that f is not one-to-one in any neighborhood of 0 (and hence not invertible
there).
[This shows that, even in the one-dimensional case n = 1, the continuity of the deriva-
tive f' at the point a is needed in the Inverse Function Theorem, even if f'(a) #0!]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81d21141-6551-4f5f-aede-0a79467fdc0f%2Fc078aa8d-de6e-41a0-968d-43a34ba88882%2F2bwnr0r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Problem 1
Consider the real-valued function of one variable f: R → R defined by
So
f(t)= =
for t= 0
t+2t² sin () for t = 0.
(a) Show that f'(0) = 1, and that the derivative f'(t) exists and is bounded in the
open interval (-1,1).
(b) Show that f'(t) is not continuous at t = 0.
(c) Show that f is not one-to-one in any neighborhood of 0 (and hence not invertible
there).
[This shows that, even in the one-dimensional case n = 1, the continuity of the deriva-
tive f' at the point a is needed in the Inverse Function Theorem, even if f'(a) #0!]
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