Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: | (52) |
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Problem 1. Let and be continuous functions defined for all , and assume that . Define a new function by
Solution. (Graded by Vicentiu Tipu (red ink) and partially regraded by Dror Bar-Natan (blue/black ink)) Let be bigger than 0, and let be a real number.
Alternative solution for the case: Clearly, by the continuity of at ,
This was not the intended solution -- it's easier than intended and doesn't indicate any understanding of arguments, but the formulation of the question doesn't rule it out, so I had to regrade many exams and give this solution full credit.
Problem 2. We say that a function is locally bounded on some interval if for every there is an so that is bounded on . Prove that if a function (continuous or not) is locally bounded on a closed interval then it is bounded (in the ordinary sense) on that interval.
Hint. Consider the set is bounded on and think about P13.
Solution. (Graded by Cristian Ivanescu) Let is bounded on as suggested in the hint. is certainly bounded on , so , so is non-empty. Every element of is in so is smaller than , so is bounded by . By P13 has a least upper bound; call it ; clearly . As is locally bounded, there is some so that is bounded on (WLOG, ) and hence on , so , so .
Assume by contradiction that . Then as is locally bounded, for some the function is bounded by some number on (WLOG, and ). As is a least upper bound, there is some with , and then is bounded by some other number on . As , it follows that is bounded by on and so , contradicting the fact that is an upper bound of . Hence it isn't true that , so therefore .
Finally, using the fact that is locally bounded, for some the function is bounded by some number on (WLOG, ). As is a least upper bound, there is some with , and then is bounded by some other number on . As , it follows that is bounded by on , so is bounded on which is what we wanted to prove.
Problem 3.
Solution. (Graded by Julian C.-N. Hung)
Problem 4. Draw a detailed graph of the function
Solution. (Graded by Vicentiu Tipu) is defined whenever , so for all . If then so . Near ,
The results. 89 students took the exam; the average grade was 75.1, the median was 76 and the standard deviation was 17.58.