Chapter 11 – Page 2

How Do You Find Where the Slope of the Tangent Line is Some Value?

A few of you had trouble with the problems referring to slopes of tangent lines. Remember, the slope of a tangent line to a function is given by the derivative. So by setting the derivative equal to the desired slope value, we can solve for the x value.

Problem 1 For the function $latex f(x)=3{{x}^{2}}-4x+1$.

1. Find where the tangent line is horizontal.
2. Find where the tangent line’s slope is equal to -1.

Problem 2 For the function $latex f(x)=2{{x}^{2}}-5x+7$.

1. Find where the tangent line is horizontal.
2. Find where the tangent line’s slope is equal to 1.

Problem 3 For the function $latex f(x)=2{{x}^{3}}-11{{x}^{2}}-8x+1$.

1. Find where the tangent line is horizontal.

In this last example, you need to factor the quadratic to find the two places where the tangent line is horizontal (has a slope of zero). In some instances, you might need to use the quadratic formula to solve the equation.

How Do You Find the Instantaneous Rate of Change?

The instantaneous rate of change is calculated to find how fast one quantity changes with respect to another.

The instantaneous rate of change of f (x)with respect to x at x = a is

$latex \displaystyle \begin{matrix}
\text{Instantaneous rate of change of }f\text{ } \\
\text{with respect to }x\text{ at }x=a \\
\end{matrix}=\underset{h\,\,\to 0}{\mathop{\lim }}\,\frac{f(a+h)-f(a)}{h}$

To apply this definition, you need to identify the point a at which the rate is to be calculated. Then the function values f (a) and f (a+h) are calculated and simplified. Finally, these are substituted into the limit so that it evaluated.

Example 1 Find the instantaneous rate of change of $latex \displaystyle f(x)=4{{x}^{2}}+2x-1$ at $latex \displaystyle x=1$.

Solution Start by calculating the two function values.

Once you have the function values, substitute them into the definition for instantaneous rate of change.

Example 2 Find the instantaneous rate of change of $latex \displaystyle f(x)={x}^{2}+6x$ at $latex \displaystyle x=2$.

Solution The function values are

Now put these into the limit definition of instantaneous rate of change.

How Do You Find The Instantaneous Rate From A Table?

The instantaneous rate of change is calculated using the limit

For many functions like polynomials, this limit may be calculated algebraically. When this limit cannot be computed algebraically or is very difficult to compute algebraically, we can use a table to estimate the limit. The problems below illustrate the table for the limit.

Problem 1 Estimate the instantaneous rate of change of f with respect to x at x = 2 if

$latex \displaystyle f(x)={{x}^{\ln (x)}}$

Solution In this problem, a = 2. We need to evaluate

$latex \displaystyle \begin{matrix}
\text{Instantaneous rate of change of }f\text{ } \\
\text{with respect to }x\text{ at }x=2 \\
\end{matrix}=\underset{h\,\,\to 0}{\mathop{\lim }}\,\frac{f(2+h)-f(2)}{h}$

Since the values in the table are shown to three decimal places, we can estimate the rate to two decimal places. In the last two columns, the difference quotient rounds to 1.12 so the rate is approximately 1.12.

Problem 2 Estimate the instantaneous rate of change of f with respect to x at x = 3 if

$latex \displaystyle f(x)={{x}^{\ln (x)}}$

Solution In this problem, a = 3. We need to evaluate

$latex \displaystyle \begin{matrix}
\text{Instantaneous rate of change of }f\text{ } \\
\text{with respect to }x\text{ at }x=3 \\
\end{matrix}=\underset{h\,\,\to 0}{\mathop{\lim }}\,\frac{f(3+h)-f(3)}{h}$

The table shows most values to 6 decimal places. In the last two columns, the values both round to 2.44864.

How Do You Maximize Revenue Using Elasticity?

There are several ways we can define elasticity E. Each indicates how the quantity demanded changes as the price is changed. In the examples below, we’ll utilize elasticity defined as

$latex \displaystyle \text{E}\approx \frac{P}{Q}\,\frac{dQ}{dP}$

With this definition, the revenue is maximized when E = -1.

Example 1 Suppose the demand function is defined by

$latex \displaystyle Q=30-\frac{P}{5}$

where Q units are demanded at a price of P dollars each.

a. Find an expression for the elasticity in terms of P.

b. Find the quantity at which the revenue is maximized.

Example 2 Suppose the demand function is defined by

$latex \displaystyle Q=40-\frac{P}{4}$

where Q units are demanded at a price of P dollars each.

a. Find an expression for the elasticity in terms of P.

b. Find the quantity at which the revenue is maximized.

Example 3 Suppose the demand function is defined by

$latex \displaystyle Q=39300-7{{P}^{2}}$

where Q units are demanded at a price of P dollars each.

a. Find an expression for the elasticity in terms of P.

b. Find the quantity at which the revenue is maximized.

How Do You Calculate A Rate From A Function?

In many business problems, we are interested in knowing the rate at which some quantity is changing. If this quantity is modeled by a function, the rate is modeled by the derivative of the function.

In the problem below, we are given a model of revenue for Verizon, a national wireless carrier. To find the rate, we’ll need to take the derivative of the revenue function.

From 2006 through 2011, Verizon Wireless grew steadily. As the number of connections increased, so did the revenue from those connections.A cubic model for the revenue over this period is

where c is the number of connections in millions.

At what rate is the revenue increasing when there are 80,000,000 connections?

Questions that ask about rate or refer to “marginal” are questions about the derivative. In this case, it is referring to the value of the derivative at c = 80. The derivative of the revenue function is

The value of the derivative at 80 million connections is

To find the units on this rate, recall that the derivative is also the slope of the tangent line at c = 80. The slope of this line has vertical units of billions of dollars and horizontal units of millions of connections. This may be simplified to

Since a billion divided by a million is one thousand. So means that an additional connection at this level results in an increase in revenue of 0.31265 thouusand dollars or $321.65.