Slant Asymptotes

We have already seen some things about rational functions

and their behavior at infinity. The behavior of this rational function at infinity is based on comparisons of the leading terms in p(x) and q(x). Specifically,

• If the exponent of the leading term of q(x) is greater than the exponent of the leading term of p(x) the function approaches 0 as a horizontal asymptote.
• If the two leading terms have the same exponent the function will approach a horizontal asymptote y = c where c is not 0.
• If the exponent of the leading term in p(x) is greater than the exponent of the leading term of q(x) the function blows up as x gets large.

Today we are going to add one more possibility that happens when the exponent of the leading term of p(x) is exactly one greater than the exponent of the leading term of q(x). In that case we will see that in the limit of large x the rational function approaches a slant asymptote whose equation is y = m x + b. More concretely, what we will see is that

To determine what m and b are, we first examine the derivative and guess that

or

Once we know what m is, we can compute b by

Example

1) Since the denominator is positive for all x, there are no vertical asymptotes and the function is defined everywhere.

2) Since the numerator has a leading term that has a degree one higher than the leading term in the denominator, this function will have a slant asymptote. To compute the equation(s) of the slant asymptote, we begin by computing the limit of the derivative as x gets large negative or positive.

The slant asymptote has the form y = 2 x + b. To determine the value of b we compute the limits

3) As we have already seen, the first derivative is

The critical points occur where the numerator is 0

Thus the only critical point is at x = 0. The first derivative is positive for all x.

4) The value of the function at the critical point is

5) The second derivative is

There are three possible inflection points: x = 0, , and . The second derivative is positive for and and negative everywhere else.

6) Here is the graph of both the function and the slant asymptote together.

Optimization Problems

An example of an applied maximum/minimum problem

Here is a problem we probably all know how to solve, if only through intuition. Consider a line in the plane, say y = 2 - 2 x. What point on that line is closest to the origin?

The intuitive answer is that if you draw a line from the origin and make that line perpendicular to the line y = 2 - 2 x, that line will cross the graph at the point closest to the origin. Keeping that in mind, let us see how to find that point by calculus methods.

To start with, let's pick any point (x,y) on the graph. What is the distance from that point to the origin?

This is the quantity that we want to minimize. A complication here is that the function appears to be a function of two variables, x and y. The methods we have developed to help us find minima only work for functions of one variable.

The resolution to this problem is to realize that x and y can not both vary independently. If we pick any point (x,y) on the line and vary its x value, the y value is constrained to vary in a specific way. We express this by writing

y(x) = 2 - 2 x

and subsequently

This simplifies the problem to the point where we can apply calculus tools. The problem now is to find the value or values of x that make the function

reach its minimum. Before we compute that minimum value, lets get a sense for this function by making a plot.

This curve has a clear minimum. All we have to do now is to find it. The minimum corresponds to a critical point of the function, that is, a point x for which . To find that critical point we have to compute the derivative.

Clearly, this equals 0 when

-4 + 5 x = 0

or

This looks right. To confirm that this is really a minimum, we could apply any one of the three tests we saw earlier. The simplest test to apply in this case is to compare the value of D(x) at 4/5 to its value at some nearby values of x.

x
	1
	0.89442719099991586
1	1

The general pattern for max/min problems

All of the applied min/max examples we are going to see will have these features.

• The quantity to be maximized or minimized is easy to write down as a function of several variables.
• Those variables can not all vary independently of each other. Some sort of constraint relation ties the variables together.
• We can use the constraint to reduce the number of variables down to one.
• We then solve the problem of maximizing or minimizing a function of one variable.
• To find extreme values, we examine the derivative to find critical points. We use some technique to determine which critical points correspond to the maxima or minima that we seek.

A second example

We want to construct a cylindrical can with volume 1000 cubic centimeters. The material used to the make the top and bottom of the can is twice as expensive as the material used to construct the side. Find the dimensions of the can that minimizes the total cost.

Given a cylindrical can with base radius r and height h, the cost of the can is proportional to

To eliminate one of the two variables here we need a constraint equation. In this case, the constraint is that the volume must be 1000:

or

Thus, as a function of r alone the cost is

We are now ready to look for a value of r that will minimize the cost. We start by computing the first derivative and finding any critical points.

Setting this equal to 0 and solving for r gives

Next, we need to confirm that this corresponds to a minimum. The second derivative is the most straightforward test to apply in this case. The second derivative is

All we need to note here is that at the critical point is clearly positive, indicating a minimum at the critical point. Thus we conclude that the can having minimal cost has dimensions

Homework

Section 4.5: 41, 60, and 67

Section 4.7: 7, 8, 21, 24, 35, 40, 54, and 59