124quiz2solutions

# 124quiz2solutions - Math 124 Quiz 2 Sept 8 2006 Name 1 Show...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 124 Quiz 2 Sept. 8, 2006 Name: 1. Show, using the rigorous definition of continuity, that the following piece-wise function is not continuous at x = 0: g ( x ) = x | x | , x 6 = 0; , x = 0 Solution: Note that if x < 0 then x | x | = x- x =- 1 and if x > 0 then x | x | = x x = 1. Hence, lim x →- g ( x ) = lim x →- x | x | = lim x →- (- 1) =- 1 and lim x → + g ( x ) = lim x → + x | x | = lim x → + (1) = 1 This shows that lim x → g ( x ) does not exist, since the limit from the left does not equal the limit from the right. Now, if g were to be continuous at 0, it would be true that lim x → g ( x ) = g (0). However, since the limit on the left doesn’t even exist, this last equation has no hope of being true. We conclude that g is not continuous at 0. 2. Show mathematically that there is a number c with 0 ≤ c ≤ 1 such that f ( c ) = 0, where f ( x ) = 2 x- 1 x . Solution: This is a case for the famous Intermediate Value Theorem. First note that f is continuous. Sinceis continuous....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

124quiz2solutions - Math 124 Quiz 2 Sept 8 2006 Name 1 Show...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online