Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: | (136) |
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**Problem 1. ** Prove that there is a real number so that

**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. **

- Define in precise terms `` is differentiable at ''.
- Let

**Solution. **

- A function is said to be differentiable at a point if the
limit
- According to the definition of differentiability, we consider the
limit

**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. **

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

**Solution. **

- See Spivak chapter 11.
- 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 .

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

The generation of this document was assisted by
L^{A}TEX2`HTML`.

Dror Bar-Natan 2002-12-09