1. 30 points Describe as fully as possible the behavior of the function
Are there any vertical, horizontal, or slant asymptotes? What is the behavior of the function near those asymptotes? Find all of the critical points and determine where the function is increasing and decreasing. Find inflection points and describe the concavity of the function. Finally, sketch the graph.
Solution The function has a vertical asymptote at x = 0. For x near 0 the numerator is approximately 1, so the behavior of the function near the asymptote is determined by the sign of the denominator. We have
Since the power in the denominator is larger than the power in the numerator, this function will go to 0 in the limit as x gets large.
By the quotient rule, the derivative of the function is
In addition to the x = 0 we have already noted, there are two critical points at 
. For 
the first derivative is negative. Outside that range it is positive. This shows that 
is a maximum point and that 
is a minimum point.
The second derivative of the function is
There are inflection points at 
. The second derivative is negative in the ranges 
and 
and positive in the ranges 
and 
.
Here is the graph.
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2. 25 points Show that among all rectangles with perimeter P the square has the largest area.
Solution A rectangle with width w and height h has area
A(w, h) = w h
and perimeter
2 w + 2 h = P
Using the second equation as a constraint to eliminate h from the area function gives
Our job is to maximize this area function. This function has first derivative
There is a single critical point at w = P/4. The second derivative test shows that this is a maximum:
With w = P/4 we have that h = P/2 - P/4 = P/4 and the rectangle is a square.
3. 20 points A fixed point of a function f(x) is a value of x for which f(x) = x. Prove that if a function f(x) is differentiable for all x and has two fixed points then for at least one value of x we must have 
Solution Let a and b be two fixed points of the function f(x). Since f(x) is differentiable for all x, we can apply the Mean Value Theorem to f(x) in the interval [a,b]. The MVT says that there is a c in [a,b] such that
Thus there is at least one x at which 
4. 20 points Use Newton's method to compute an approximation for 
accurate to four decimal places.
Solution The first thing we need is a function f(x) that has 
as one of its roots. An obvious candidate is
The Newton formula applied to this function gives
From a starting guess of x0 = 3 we get the following sequence of points:
| n | xn |
|---|---|
| 0 | 3 |
| 1 | 3.111111111111111 |
| 2 | 3.107237339380196 |
| 3 | 3.107232505961377 |
Accurate to four decimal places we have that 
. The actual value is
