How Do You Find Quantities Related to Revenue and Cost?

In this problem, you were given revenue and cost functions like \displaystyle R(x)=-{{x}^{2}}+8x and \displaystyle C(x)=2x+5. In the parts of the problem you were asked to a) identify the correct graph, b) find the break-even quantity, c) the maximum revenue, and d) the maximum revenue. In class yesterday, the students worked in groups to solve this problem. Here is the work they put on the board.

Parts a and b

Parts c and d

Look pretty good but on part d, they should write the profit at x = 4 as P(4). The key in thee part is to use \displaystyle x=-\frac{b}{2a} on the quadratic revenue and profit functions to find the quantities at which the maximums are achieved. Since the parts asked you to find the maximum revenue and profit, you need to put the quantities into the functions.

How Do You Compute a Limit Using a Table?

In each of the next two examples, the value of the limit is the same as the value of the function at the point it approaches. This is typically the case for any polynomial.

For functions that are not polynomials, a table i often in order to evaluate the limit. In each of the following two examples, the output of the function grow more positive or more negative. This means the different limits do not exist.

However, a very similar looking fraction may also lead to a limit that does exist. In the next two examples, the limits do exist even though the functions are undefined at the point the x value is approaching.

How Do You Compute a Limit From a Graph?

In the next several examples, a graph is used to evaluate one-sided limits from the right and left, the corresponding two-sided limit, and the value of the function. The key is to find the y values as the x values approach some value from the left and the right…if those match, the two sided limit is the y value that it matches at.

Does anyone see any mistakes in the examples above?

How Do You Calculate a Limit Algebraically?

You can recognize the limits by what happens when you substitute the value x approaches into the expression. If it gives 0/0, there is algebra that you can do to find the exact value of the limit.

In the first two examples, the expression may be factored and simplified…then you can substitute the value for x.

\underset{x\to 3}{\mathop{lim }}\frac{{{x}^{2}}-5x+6}{x-3}

\underset{x\to -1}{\mathop{lim }}\frac{{{x}^{2}}-x-2}{x+1}

In the next two examples, the fractions in the numerator must be combined before the fraction may be simplified.

\underset{x\to 0}{\mathop{lim }}\frac{\frac{1}{x-6}+\frac{1}{6}}{x}

\underset{x\to 0}{\mathop{lim }}\frac{\frac{1}{4}-\frac{1}{x+4}}{x}

The next two examples are designed to throw you off. When you substitute the value into the expression, you do not get 0/0. This means you need to use a table or graph to get the limit.

\underset{x\to -5}{\mathop{lim }}\frac{1}{{{\left( x+5 \right)}^{2}}}

\underset{x\to 1}{\mathop{lim }}\frac{x}{{{\left( x-1 \right)}^{2}}}

The next two examples show how to rationalize the numerator to do a limit.

\underset{x\to 1}{\mathop{lim }}\frac{\sqrt{x}-1}{x-1}

This example may be done two different ways as the next two boards demonstrate.

\underset{x\to 4}{\mathop{lim }}\frac{\sqrt{x}-2}{x-4}

How Do You Compute the Limit of a Difference Quotient?

The last part of Section 10.3 asks you to compute several different difference quotients. Some of the problems ask you to compute

\underset{h \to 0}{\mathop{lim }},\frac{f(a+h)-f(a)}{h}

where f(x) and a are given to you in the problem. Here are a few examples from the board.

f(x)=4x+3 and a=1

The board above contains a mistake…do you see where this group made a mistake?

f(x)={{x}^{2}}-4 and a=1

f(x)={{x}^{2}}-1 and a=2

 

In the examples below, you are asked to compute a difference quotient containing x instead of a.

Compute \underset{h \to 0}{\mathop{lim }},\frac{f(x+h)-f(x)}{h} where f(x)={{x}^{2}}+2x

 

Compute \underset{h \to 0}{\mathop{lim }},\frac{f(x+h)-f(x)}{h} where f(x)={{x}^{2}}-x

 By the way, the correct solution to the first problem is below.

In this original calculation, f(a+h) and f(a) where switched.