Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: (136) Next: Integration Previous: Term Exam 2

# Solution of Term Exam 2

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Problem 1. Prove that there is a real number so that

If your proof uses the intermediate value theorem, state it clearly and prove that it follows from the postulate P13.

Solution. As a composition/sum/quotient of continuous functions, the left hand side is a continuous function of . The term is bounded by 157 and hence the large behaviour of the left hand side is dominated by that of . Thus for large negative the left hand side goes to and for large positive it goes to . Thus by the intermediate value theorem the left hand side must attain the value for some .

Our proof does use the intermediate value theorem, and hence its statement and proof should be reproduced. See Spivak's chapter 8.

Problem 2.

1. Define in precise terms  is differentiable at ''.
2. Let

Is differentiable at 0? If you think it is, prove your assertion and compute . Otherwise prove that it isn't.

Solution.

1. A function is said to be differentiable at a point if the limit

exists.
2. According to the definition of differentiability, we consider the limit

We claim that this limit is 0 and hence exists and is equal to 0. Indeed, Let be any positive number and set . Now if satisfies is rational then and if satisfies is irrational then , so in general implies . Thus as asserted above.

Problem 3. Calculate in each of the following cases. Your answer may be in terms of , of , or of both, but reduce it algebraically to a reasonably simple form. You do not need to specify the domain of definition.

 (a) (c) (b) (d)

Solution.

(a)
Differentiating both sides with respect to we get and hence .

(b)
Using the rule for differentiating a quotient, then the chain rule and then simplifying a bit, we get

(c)
Differentiating both sides with respect to we get and hence .

(d)
Using the chain rule, .

Problem 4.

1. Prove that if on some interval then is increasing on that interval.
2. Sketch the graph of the function .

Solution.

1. See Spivak chapter 11.
2. is not defined; . The only solution to is , so the point is on the graph. ; this is positive when and when and negative when , so is increasing when and when and decreasing when . The derivative is 0 only at ; right before, the function is decreasing and right after it is increasing. So is a local max and we can compute . Finally, and near our graph is very close to , so we arrive at the following graph:

Problem 5. Write a formula for in terms of , and . Under what conditions does your formula hold?

Solution. From class material we knot that if is continuous and near , differentiable at , and then . Using this we get

In the last chain of equalities we've used the chain rule, for which, in addition to what we already have, we need to know that is continuous around and differentiable at and the rule for differentiating a quotient, for which we need nothing new. Hence the full list of conditions needed for aour formula to hold is:
• is near .
• is differentiable around .
• .
• is twice differentiable at .

an alternative solution:

The results. 86 students took the exam; the average grade is 70.76, the median is 72 and the standard deviation is 18.35.

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Dror Bar-Natan 2002-12-09