This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2 f ( x) = ⎨ྏ
⎩ྏ x − 2 2 ≤ x < 5
⎧ྏ2 − x − 5 ≤ x < 2 g ( x) = ⎨ྏ
x−2 2≤ x <5
⎩ྏ f ( x) is continuous over the interval (− ∞,0), (0,1), (1, ∞) since each of the interval is Solution: On the interval, we know that the limit exits as and x → −5+ and x → 5− . These are from the boundary conditions of f ( x) . So we need to test whether the made up of a polynomial function. Now, the lim f ( x) = lim x + 2 = 2
x →0 − x →0 − limit exits at x=2 lim+ f ( x) = lim+ 2 x 2 = 0 so the function is x→0 lim f ( x) = lim− 3 − x = 1 x →0 x →2 − discontinuous at x = 0 . Also lim 2 x 2 = 2 and x →2 and lim f ( x) = lim+ x − 2 = 0 x →2 + x →1− x →2 At x=2, the limit DNE, so it is not continuous. Therefor
f ( x) is continuous on the intervals [−5,2) ∪ (2,5] For g ( x) lim 2 − x = 1, so the function is discontinuous x→1+ at x = 1 . Since, f (1) = 2 the...
View
Full
Document
This document was uploaded on 02/27/2014 for the course M 408c at University of Texas at Austin.
 Winter '06
 McAdam
 Continuity, Asymptotes, Differential Calculus, Limits

Click to edit the document details