Dror Bar-Natan: Classes: 2003-04: Math 157 - Analysis I: (52) Next: Class Notes for Tuesday December 2, 2003 Previous: Term Exam 2

# Solution of Term Exam 2

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Problem 1. Let and be continuous functions defined for all , and assume that . Define a new function by

Is continuous for all ? Prove or give a counterexample.

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.

• If use the continuity of at to find so that . Set and then if then so is continuous at .

• If use the continuity of at to find so that . Set and then if then so is continuous at .

• If use the continuity of at 0 to find so that and use the continuity of at 0 to find so that . Set . Now if then also and hence and if then also hence . Combining the last two assertions we find that , so is continuous at .

Alternative solution for the case: Clearly, by the continuity of at ,

and by the continuity of at ,

Thus the two one-sided limits of exist and are equal (to ). By an exercise we did earlier, this implies that also exists and also equals , so is continuous 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.

1. Prove that if a function satisfies on then for some constant .
2. A certain function was differentiated twice, and to everybody's surprise, the result was back the function again, except with the sign reversed: . It was also found that . Set and compute , and (making sure that you explain every step of your computation).

Solution. (Graded by Julian C.-N. Hung)

1. Let be two real numbers. By the mean value theorem there is some so that

so so . As and were arbitrary, it follows that is the constant function.

2. , so for some constant . Now . But is a constant, so is also .

Problem 4. Draw a detailed graph of the function

Your drawing must clearly indicate the domain of definition of , all intersections of the graph of with the axes, the behaviour of far out near and near the boundaries of its domain of definition, the regions on which it is increasing or decreasing, the regions on which it is convex or concave and all local and global minima and maxima of .

Solution. (Graded by Vicentiu Tipu) is defined whenever , so for all . If then so . Near ,

Near the numerator of goes to and the denominator is positive but approaches 0. So . The derivative is . This is positive for and negative for and for , so is increasing for and decreasing for and for . At the derivative is 0; before the function is increasing and after it is decreasing, so has a local maximum at . There are no other points in which , so there are no other local maxima and minima. By inspecting the value of at , and comparing it with the behaviour of at and near 0, we find that is a global maximum. Finally, differentiating a second time we get . We see that for and for . So is convex on and concave on and on . Summarizing, the graph looks as follows:

The results. 89 students took the exam; the average grade was 75.1, the median was 76 and the standard deviation was 17.58.

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Dror Bar-Natan 2004-10-18