Category: Chapter 10

How Do You Calculate a Limit at Infinity?

You may have notice that there are two ways to use algebra to compute limits at infinity. Let’s look at two possible strategies for computing a limit at infinity for a rational function, $latex \underset{x \to – \infty }{\mathop{lim }},\frac{2{{x}^{3}}+x}{3{{x}^{2}}+1}$.

In the board above, I divided each term in the rational expression by x to the highest power that appears anywhere in the expression. The denominator get very small as the top approaches 2, so the entire expression must grow very big. Since the bottom is also negative, the expression grows big and is negative. Now let’s change that strategy a bit.

In the board above, I divide by x to the largest power that appears in the denominator. The bottom then gets closer and closer to 3. But the top contains 2x which grow more and more negative as x approaches $latex – \infty$. So the entire expression grows more and more negative.

Either strategy gives the correct answer as long as you interpret what is going on in the numerator and denominator correctly.

How Do I Deal With Variables Other Than x in Linear Functions?

When working with functions with unusual variable names and scaling, students often run into two problems.

  • trouble with what each variable represents
  • trouble with the scaling of the variables

Since we rarely use x in Finite Math to represent the independent variable, we need to get used to these names. Typically, the variables are aligned or scaled. This means we need to pay particular care to interpreting answers.

In one type of the problem, you are given a demand function $latex \displaystyle p=D(q)=32-1.25q$ where p is the price in dollars and q is the quantity of watches demanded in hundreds. In part b, you were asked to find the price when the demand is 800 watches. This quantity corresponds to $latex \displaystyle q=8$ and gives a price of

$latex \displaystyle D(8)=32-1.25(8)=22$

Notice how the scaling on the variable works…800 watches is input as 8 hundred. In part c, you need to find the demand when the price is 27 dollars. Now we know the output from the function and we need to find the corresponding input,

$latex \displaystyle 32-1.25q=27$

When we solve this for q we get $latex \displaystyle q=4$. In other words, 4 hundred watches are demanded when the price is 27 dollars. Since the units on the answer blank is watches, you will need to type 400 in the answer blank. I think many of you were not taking the units into account on this problem. Units are incredibly important in the real world where you are dealing with millions of dollars. In those cases misinterpreting 2,000,000 dollars and 2 million dollars may cost you your job.

How Do You Find Break-Even Points?

If you are looking for where the break-even points are, you must determine the quantity for which

revenue = cost

Alternately, you may find the profit by calculating

profit = revenue – cost

The problem below demonstrate these strategies starting from a demand function and a cost function. To apply either of the relationships above, you need to form the revenue function from

revenue = (price)(quantity)

where the price is given by the demand function and Q represents the quantity.

Problem The demand function for Q units of a product is given by

$latex \displaystyle D\left( Q \right)=16-1.25Q$

The cost function is given by the function

$latex \displaystyle C\left( Q \right)=2Q+15$

a. Find the revenue function R(Q).

b. Find the break-even point(s)?

c. On a graph of R(Q) and  C(Q), where do the break-even points lie?

d. Find the profit function P(Q).

e. Where do the break-even points lie on the graph of P(Q)?

Solution 1 To find the break-even point, this group of students set R(Q) = C(Q). This results in a quadratic equation. They moved all terms to one side and used the quadratic formula to find the quantities at which the revenue is equal to the cost.

Solution 2 This group of students found the profit function P(Q) first. Then they set it equal to zero to find the break-even points. Like the first solution, they also needed to use the quadratic formula.

Both techniques lead to the same break-even points and are equally valid. The only thing the second solution left out was the graph of the profit function showing the break-even points at the zeros (horizontal intercepts) of the function.