MATH 100 Lecture Notes - Lecture 11: Classification Of Discontinuities, California State Route 1, In C

• School: UBC Okanagan, UBC OK
• Department: Mathematics
• Course: MATH 100
• Professor: Majid Jaberipour

Document Summary

5 a " fgu , =/ t. 5am ios ( i a cos in - I ) t b i u ] flu ) =n f flu ) Contino us all value between a and p. I a b elastance of at least one root of interval. If discontinuity because of limit does not exist we can "t have a root somewhere in ( a. b) do anything to fix that. If f ca ) f (b) co , Then we have at least one root fix the problem. Find the value of function: g . I u -42 ) is continuous somewhere between. 4 is fth ) must he y so f ( - 11. =f ( u i is continuous fl. 11=2 i. flu , has a root in. 2 has a solution in f- 10,0 ] finding parameters so function is continuous at. I denominator should not have roots t c -