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Unformatted text preview: Math 221, Lecture 7, Fall 2006, Midterm 1. Try to do first the problems that look simplest to you. You dont have to follow the order. Avoid spending too much time on one question with others not done yet. Good luck! 1. (15) Consider the function f ( x ) given by f ( x ) = x 2 , x <- 2 4 ,- 2 x < 1 5 , x = 1 x + 3 , 1 < x 2- x + 7 , x > 2 Sketch the graph of f. At which points a the limit lim x a f ( x ) does not exist? At which points f ( x ) is not continuous? At which points f ( x ) is not differentiable? Justify your answers. The limit lim x a f ( x ) exists at every point a, since right and left limits exist and coincide at all points. Notice that at x = 1 the limit exists and is equal to 4 , even though f (1) = 5 . By definition of the limit, we look at values of f ( x ) as x approaches 1 , not at x = 1 . The function f is not continuous at x = 1 , since the value of the limit at this point is not equal to the value of the function: lim x 1 f ( x ) = 4 6 = 5 = f (1) . At the rest of the points f is continuous by continuity of linear functions and x 2 that was discussed in class. f ( x ) is not differentiable at x = 1 , since f is not even continuous at this point. In addition, f is not differentiable at x = 2 and x =- 2 . Indeed, at x = 2 we have ( f (2 + h )- f (2)) /h = (- 2- h + 6- 4) /h =- 1 if h > 0 and ( f (2 + h )- f (2)) /h = (2 + h + 2- 4) /h = 1 if h < . Therefore, the limit lim h f (2 + h )- f (2) h does not exist, as the limits from the left and from the right are not equal. Similarly, at x =- 2 ( f (- 2 +...
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This note was uploaded on 03/27/2008 for the course MATH 211 taught by Professor Onlineresources during the Spring '06 term at Wisconsin.
- Spring '06