A tangent line to a function is a line that looks most like the function at a point. In common terms, it just grazes the function.

To find its equation, we need to locate the point where the two meet as well as the slope of the function at that point. Then we can use the slope-intercept form or point-slope form of a line to get the equation.

Problem Find the equation of the tangent line to

atx = 3

Solution Since this problem is asking for the equation of a line, let’s start with the point-slope form

This requires a point (x_{1}, y_{1}) and slope m. We’ll use the function to get the point and the derivative to get the slope of the tangent line.

Find the point: We are given a pointx = 3. To find the corresponding y value, put the x value into the function

Find the slope of the tangent line: We need h′(3) to get the slope of the tangent line. We’ll use the Power Rule to take the derivative,

The slope of the tangent is

Write the equation of the tangent line: Putting the point (3, 10) and the slope 9 into the line yields

If you are asked to write this in slope-intercept form, you’ll need to solve this for y to give

If you graph h(x) and the tangent line together, it should be obvious that your tangent line is correct (ie. tangent).

A tangent line to a function is a line that looks most like the function at a point. In common terms, it just grazes the function.

To find its equation, we need to locate the point where the two meet as well as the slope of the function at that point. Then we can use the slope-intercept form or point-slope form of a line to get the equation.

Find the equation of the tangent line to

at x = 3.

Since this problem is asking for the equation of a line, let’s start with the point-slope form

This requires a point (x_{1}, y_{1}) and slope m. We’ll use the function to get the point and the derivative to get the slope of the tangent line.

Find the point: We are given a point x = 3. To find the corresponding y value, put the x value into the function

Find the slope of the tangent line: We need h′(3) to get the slope of the tangent line. We’ll use the Power Rule to take the derivative,

The slope of the tangent is

Write the equation of the tangent line: Putting the point (3, 10) and the slope 9 into the line yields

If you are asked to write this in slope-intercept form, you’ll need to solve this for y to give

If you graph h(x) and the tangent line together, it should be obvious that your tangent line is correct (ie. tangent).

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.

In each problem below, the average cost function by dividing the cost function by the variable representing the quantity. For a cost function C(Q), the average cost function is

Find and interpret the marginal average cost when 20 units are produced.

This means that each of the 20 units costs an average of .1386 hundred dollars or $13.86.

In this board they have used the fact that dividing by Q is the same as multiplying by 1/Q.

Although it is OK to leave the derivative unsimplified, they need to put in 20. So it is best to do some algebra before putting in the value. Since -0.006 is the slope of the tangent line on the average cost function, the units on it is hundreds of dollars per unit per unit:

$latex \displaystyle \frac{-0.006}{1}\frac{\frac{\text{hundreds of dollars}}{\text{unit}}}{\text{unit}}$

This means that if production is increased by 1 unit, the average cost will drop by 0.006 hundred dollars per unit.

Problem 2 Suppose the total cost function for a product is

Find and interpret the marginal average cost when 10 units are produced.

This value tells us that if production is increased by 1 unit, the average cost will drop by 0.3472 thousand dollars per unit or $347.2 per unit. Had they rounded one more decimal place, we would have had this number to the nearest penny.

Problem 3 Suppose the total cost (in thousands of dollars) to produce Q units is

a. Find the the average cost of producing 40 units.

b. Find the average cost function $latex \displaystyle \overline{TC}\left( Q \right)$.

c. Find the marginal average cost function $latex \displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$.

d. Find and interpret $latex \displaystyle \overline{TC}{{\,}^{\prime }}\left( 40 \right)$.

The marginal average cost is simply the slope of the tangent line to the average cost$latex \displaystyle \overline{TC}{{\,}^{\prime }}\left( Q \right)$. The slope has vertical units of thousands of dollars per unit and horizontal units of units. So the rate has units of thousands of dollars per unit per unit. This means that if the production were to increase by one, the average cost would drop by 0.0007701 thousand dollars per unit or 0.7701 dollars per unit. This means that the average cost is decreasing…probably a good thing for the bottom line.

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

Find where the tangent line is horizontal.

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

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

Find where the tangent line is horizontal.

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

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

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.