B Midterm Practice Problem sol - PRACTICE PROBLEMS FOR MIDTERM SOLUTIONS The test will be given on Wednesday April 28 It will cover Sections 17-20 and

# B Midterm Practice Problem sol - PRACTICE PROBLEMS FOR...

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PRACTICE PROBLEMS FOR MIDTERM – SOLUTIONS The test will be given on Wednesday, April 28 . It will cover Sections 17-20 and 28-29 (everything covered in this quarter, up to and including Mean Value Theorem). In preparing for the test, you can practice solving the problems from the list below. In addition, take a look at the homework assignments 1-3 ( at least one problem on the midterm will come from the homework ), at the examples given in the textbook, and the the examples discussed in class. 1. Suppose the functions f , g , and h are defined on the interval ( a, b ), and g ( x ) f ( x ) h ( x ) for every x ( a, b ). Suppose, furthermore, that g and h are continuous at c ( a, b ), and g ( c ) = h ( c ). Prove that f is continuous at c . We need to show that, if ( x n ) is a sequence of elements of ( a, b ), satis- fying lim x n = c , then lim f ( x n ) = f ( c ). But g ( x n ) f ( x n ) h ( x n ), and lim g ( x n ) = g ( c ) = f ( c ), and lim h ( x n ) = h ( c ) = f ( c ). By Squeeze Theorem, lim f ( x n ) = f ( c ). 2. Prove that there exists x [0 , 2] s.t. x 5 3 x 1 = x 3 + 1. Consider f ( x ) = x 5 3 x 1 x 3 + 1. Then f is continuous on [0 , 2]. Indeed, f ( x ) = g ( x ) h ( x ), where g ( x ) = x 5 3 x 1, h ( x ) = radicalbig k ( x ), and k ( x ) = x 3 + 1. g and k are polynomials, hence they are continuous. Moreover, k ( x ) 0 for any x [0 , 2], hence h is also continuous on [0 , 2]. This implies the continuity of f = g h . Observe that f (0) = 2, and f (2) = 32 6 1 3 = 22. By Intermediate value Theorem, f ( x ) = 0 for some x (0 , 2). Therefore, x 5 3 x 1 = x 3 + 1. 3. Let f ( x ) = 1 / x . Prove that f is not uniformly continuous on (0 , 1]. Solution 1 . We know that a uniformly continuous function maps a Cauchy sequence to a Cauchy sequence. This is not the case for f . Indeed, let s n = 1 /n ( n N ). The sequence ( s n ) is convergent, hence Cauchy. Moreover, s n (0 , 1] for any n . However, f ( s n ) = n , and lim f ( s n ) = + . Thus, the sequence ( f ( s n )) is not Cauchy.