410HW1 - Math 410 HW 1 Solution:s January 24, 2009 1....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 410 HW 1 Solution:s January 24, 2009 1. Consider the function f defined on the unit interval [0 , 1] by f ( x ) = if x = 0 x sin(1 /x ) if 0 < x 1. (a) Is f continuous on [0 , 1]? Solution: The function f as defined on the half-open interval (0 , 1] is obviously continuous, being the composition and product of continu- ous well-defined functions. We therefore only need to check continuity of f at x = 0. Note that- x x sin(1 /x ) x so that the squeeze theorem implies lim x + f ( x ) = 0 = f (0) . This guarantees that f is also continuous at x = 0. Thus, f is indeed continuous on [0 , 1]. (b) Does f belong to C 1 [0 , 1]? Solution: Standard calculus reveals that for x in the open interval (0 , 1), f ( x ) =- 1 x cos 1 x + sin 1 x , which is continuous. However, there is no possible way to guess the value of f (0) without doing some work. Using the definition of the derivative, f (0) = lim h + f (0 + h )- f (0) h = lim h + f ( h )- h = lim h + h sin(1 /h ) h = lim h + sin(1 /h ) , which does not exist. Hence, f does not have a derivative (much less a continuous derivative) at x = 0, so f does not belong to C 1 [0 , 1]. 1 2. Find a function g C ( R ) which satisfies g ( x ) = if x - 1 1 if x 0....
View Full Document

Page1 / 4

410HW1 - Math 410 HW 1 Solution:s January 24, 2009 1....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online