For a statement of the problems and a summary of the solutions of Exam 3, see Exam 3 Solutions.

1. To prove that if  f  is integrable on  [a, c]  and on  [c, b],  then  f  is integrable on  [a, b]  (for  a < c < b).
Comment.  See Exercise 6, Section 6.3. This is NOT the same as Theorem 6.12 on page 150; it is a kind of converse. Theorem 6.12 says that integrability on the “big” interval implies integrability on each of the smaller intervals. On the exam, you are to prove that integrability on each of the smaller intervals implies integrability on the big interval.
Exercise 6, Section 6.3, was not given explicitly as a homework problem, but the proof could be given as part of the solution of Exercise 8 in Section 6.4. It is a direct consequence of the Archimedes-Riemann Theorem applied to each of the intervals, or the “Cauchy criterion for integrability”, which is the “necessary and sufficient” form of Exercise 4 in Section 6.4.

2. To prove that if the integral of a nonnegative continuous function is zero, then the function is zero.
Comment.  This is essentially the same as to Exercise 5 in Section 6.4, which was assigned for homework. It is important to realize that this result, and the result in Exercise 5, is NOT necessarily true if the function is not continuous, and therefore continuity (or a consequence of continuity) must be used somewhere in the proof. In you give a proof which does not specifically invoke continuity, or a specific consequence of continuity, the proof is almost certainly wrong.

3. To prove the differentiability at 0 of a given function defined by one formula on the rationals, a different formula on the irrationals.
Comment.  There have been several examples of functions defined in this way, and several examples of functions with the given behavior at the origin. See, in particular, Exercise 9 in Section 3.7 (which was assigned as homework) and, especially, Problem 2 on Exam 2 (and the solution provided). The desired result here is a direct consequence of the result of Ex. 9 in Sect. 3.7 and can easily be proved directly using the same technique. Note especially that the best approach here may not be to use the sequential definition of limit but to use the epsilon-delta definition. And if you do use the sequential definition, it is still much easier to use an inequality FIRST, instead of trying to apply sequences to the two cases given in the definition of the function (as in Ex. 9 in Sect. 3.7 and in the solution to Problem 2 on Exam 2). See email of 22 Nov.

4. To prove that a certain polynomial equation has exactly two solutions.
Comment.  There are many examples like this in the exercises in Section 6.3, some of which were assigned as homework. The proof uses the Intermediate Value Theorem for continuous functions (NOT a calculator) to prove that solutions exist and uses the Mean Value Theroem for differentiable functions (or consequences thereof) to get an upper bound on the number of solutions. Note that in using the MVT here, one can either use it directly, as is suggested in the solution posted, or one can use the consequence that a function with a positive derivative is strictly increasing -- in this case, applied to the derivative of the given polynomial function. In any case, the pattern is: IVT (or consequence thereof) implies existence, MVT (or consequence thereof) implies some kind of “uniqueness”.

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