Chapter 1 Functions

§ 1.1 The Slope of a Straight Line

DefinitionExample Equations of Nonvertical Lines: A nonvertical line L has an equation of the form The number m is called the slope of L and the point (0, b) is called the y-intercept. The equation above is called the slope-intercept equation of L. For this line, m = 3 and b = -4. Nonvertical Lines

Lines – Positive SlopeEXAMPLE The following are graphs of equations of lines that have positive slopes.

Lines – Negative SlopeEXAMPLE The following are graphs of equations of lines that have negative slopes.

Interpretation of a GraphEXAMPLE A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this straight line.

Properties of the Slope of a Nonvertical Line

Properties of the Slope of a Line

Finding Slope and y -intercept of a LineEXAMPLE Find the slope and y-intercept of the line

Making Equations of LinesEXAMPLE Find an equation of the line that passes through the points (-1/2, 0) and (1, 2).

Making Equations of LinesEXAMPLE Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x.

Slope as a Rate of ChangeEXAMPLE Compute the rate of change of the function over the given intervals.

§ 1.2 The Slope of a Curve at a Point

Tangent Lines DefinitionExample Tangent Line to a Circle at a Point P: The straight line that touches the circle at just the one point P

Slope of a Curve & Tangent Lines DefinitionExample The Slope of a Curve at a Point P: The slope of the tangent line to the curve at P (Enlargements)

Slope of a GraphEXAMPLE Estimate the slope of the curve at the designated point P. The slope of a graph at a point is by definition the slope of the tangent line at that point. The figure above shows that the tangent line at P rises one unit for each unit change in x. Thus the slope of the tangent line at P is

Slope of a Curve: Rate of Change

Interpreting Slope of a GraphEXAMPLE Refer to the figure below to decide whether the following statements about the debt per capita are correct or not. Justify your answers. (a) The debt per capita rose at a faster rate in 1980 than in (b) The debt per capita was almost constant up until the mid-1970s and then rose at an almost constant rate from the mid-1970s to the mid-1980s.

Interpreting Slope of a Graph (a) The slope of the graph in 1980 is marked in red and the slope of the graph in 2000 is marked in blue, using tangent lines. It appears that the slope of the red line is the steeper of the two. Therefore, it is true that the debt per capita rose at a faster rate in 1980.

Interpreting Slope of a Graph (b) Since the graph is a straight, nearly horizontal line from 1950 until the mid-1970s, marked in red, it is therefore true that the debt per capita was almost constant until the mid-1970s. Further, since the graph is a nearly straight line from the mid-1970s to the mid-1980s, marked in blue, it is therefore true that the debt per capita rose at an almost constant rate during those years.

Equation & Slope of a Tangent LineEXAMPLE Given the slope of the graph of y = x 2 at the point (x, y) is 2x. Find the slope of the tangent line to the graph of y = x 2 at the point (-0.4, 0.16) and then write the corresponding equation of the tangent line.

§ 1.3 The Derivative

DefinitionExample Derivative: The slope formula for a function y = f (x), denoted: Given the function f (x) = x 3, the derivative is

Differentiation DefinitionExample Differentiation: The process of computing a derivative. No example will be given at this time since we do not yet know how to compute derivatives. But don’t worry, you’ll soon be able to do basic differentiation in your sleep.

Differentiation Examples These examples can be summarized by the following rule.

Differentiation ExamplesEXAMPLE Find the derivative of

Differentiation ExamplesEXAMPLE Find the slope of the curve y = x 5 at x = -2.

Equation of the Tangent Line to the Graph of y = f ( x ) at the point ( a, f ( a ))

Equation of the Tangent LineEXAMPLE Find the equation of the tangent line to the graph of f (x) = 3x at x = 4.

Leibniz Notation for Derivatives Ultimately, this notation is a better and more effective notation for working with derivatives.

Calculating Derivatives Via the Difference Quotient The Difference Quotient is

Differentiable DefinitionExample Differentiable: A function f is differentiable at x if approaches some number as h approaches zero. The function f (x) = |x| is differentiable for all values of x except x = 0 since the graph of the function has no definite slope when x = 0 (f is nondifferentiable at x = 0) but does have a definite slope (1 or -1) for every other value of x.

Limit Definition of the Derivative

Use TI89 to Graph I.Slope and tangent lines 1) Graph the function. 2) 2 nd Draw 5, then type x value or graph 2 nd calc 6, then type x value. II.Graph y and y’ at the same time 1) graph the function in y1. 2) Enter y2 = nDerive(y1, x, x), then graph: y2 = Math nDerive( Vars Yvars y1 then press, x, x) graph

§ 1.4 Limits and the Derivative

Definition of the Limit

Finding LimitsEXAMPLE SOLUTION Determine whether the limit exists. If it does, compute it. Let us make a table of values of x approaching 4 and the corresponding values of x 3 – 7. x x As x approaches 4, it appears that x 3 – 7 approaches 57. In terms of our notation,

Finding LimitsEXAMPLE SOLUTION For the following function g(x), determine whether or not exists. If so, give the limit. We can see that as x gets closer and closer to 3, the values of g(x) get closer and closer to 2. This is true for values of x to both the right and the left of 3.

Limit Theorems

Finding LimitsEXAMPLE Use the limit theorems to compute the following limit.

Limit Theorems

Finding LimitsEXAMPLE Compute the following limit.

Using Limits to Calculate a Derivative

Limit Calculation of the DerivativeEXAMPLE Using limits, apply the three-step method to compute the derivative of the following function:

Using Limits to Calculate a DerivativeEXAMPLE Use limits to compute the derivative for the function

Limits as x Increases Without BoundEXAMPLE Calculate the following limit.

§ 1.6 Some Rules for Differentiation

Rules of Differentiation

Differentiation Differentiate EXAMPLE

Differentiation Differentiate EXAMPLE

The Derivative as a Rate of Change = Slope of the tangent line at the point (a, f(a))

The Derivative as a Rate of ChangeSOLUTION Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a certain department store. Represent each statement below by an equation involving S or. EXAMPLE (a) The sales at the end of January reached $120,560 and were rising at the rate of $1500 per month. (b) At the end of March, the sales for this month dropped to $80,000 and were falling by about $200 a day (Use 1 month = 30 days). (a) Since the sales at the end of January (the first month, so x = 1) reached $120,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 120,560. Further, since the rate of change of sales (rate of change means we will use the derivative of S(x)) for the month of January is a positive $1500 per month, we have: (b) At the end of March (the third month, so x = 3), the sales dropped to $80,000. Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000. Additionally, since sales were dropping by $200 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-200 dollars) = dollars per month. Therefore, we have:

§ 1.7 More About Derivatives

Differentiating Various Independent Variables Find the first derivative. EXAMPLE

Second Derivatives Find the first and second derivatives. EXAMPLE

Second Derivatives Evaluated at a Point Compute the following. EXAMPLE

Marginal Cost

SOLUTION Let C(x) be the cost (in dollars) of manufacturing x bicycles per day in a certain factory. Assume C(50) = 5000 and Estimate the cost of manufacturing 51 bicycles per day. EXAMPLE We will first use the additional cost formula for manufacturing 1 more bicycle per day beyond the cost of producing 50 bicycles per day. We already know it costs $5000 to produce 50 bicycles per day since C(50) = So we wish to determine how much more, beyond that $5000, it costs to produce 51 bicycles. Therefore, we estimate the cost of manufacturing 51 bicycles to be $5045.

Marginal Revenue & Marginal Profit

Marginal RevenueSOLUTION Suppose the revenue from producing (and selling) x units of a product is given by R(x) = 3x – 0.01x 2 dollars. EXAMPLE (a) Since we are looking for the marginal revenue at a production level of 20, and we have an equation for R(x), we will simply find (a) Find the marginal revenue at a production level of 20. (b) Find the production levels where the revenue is $200. This is the given revenue function. This is the marginal revenue function. Evaluate the marginal revenue function at x = 20. Therefore, the marginal revenue at a production level of 20 is 2.6.

Marginal Revenue (b) To find the production levels where the revenue is $200 we need to use the revenue function and replace revenue, R(x), with 200 and then solve for x. This is the given revenue function. Replace R(x) with 200. Therefore, the production levels for which revenue is $200 are x = 100 and x = 200 units produced. CONTINUED Get everything on the left side of the equation. Multiply everything by 100. Factor. Solve for x.

§ 1.8 The Derivative as a Rate of Change

Average Rate of Change

Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2? EXAMPLE

Instantaneous Rate of Change

Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1? EXAMPLE

Average & Instantaneous Rates of Change Refer to the figure below, where f (t) is the percentage yield (interest rate) on a 3-month T-bill (U.S. Treasury bill) t years after January 1, EXAMPLE (a) What was the average rate of change in the yield from January 1, 1981 to January 1, 1985? (b) How fast was the percentage yield rising on January 1, 1989? (c) Was the percentage yield rising faster on January 1, 1980 or January 1, 1989?

Average & Instantaneous Rates of ChangeSOLUTION (a) To determine the average rate of change in the yield from January 1, 1981 to January 1, 1985, we must first determine the coordinates of the two points that correspond to the two given dates. They are (1, 14) and (5, 7). Now we use the average rate of change formula. CONTINUED Therefore, the average rate of change in the yield from January 1, 1981 to January 1, 1985 is -7/4. (b) To determine how fast the percentage yield was rising on January 1, 1989, we must determine the instantaneous rate of change of f (t) when t = 9 (corresponding to January 1, 1989). This means that we must find the slope of the tangent line to the graph of f (t) where t = 9. The tangent line is on the graph and so we need only determine any two points on the tangent line. Using the coordinates of these two points, we will calculate the slope of the tangent line and that will be the instantaneous rate of change that we seek. Notice that two of the points on the tangent line are (5, 5) and (11, 10). Using these points we will now calculate the slope of the tangent line.

Average & Instantaneous Rates of ChangeCONTINUED Therefore, the rate at which the percentage yield was rising on January 1, 1989 was 5/6. (c) To determine if the percentage yield was rising faster on January 1, 1980 or January 1, 1989, we would need to know the slopes of the tangent lines corresponding to t = 0 (January 1, 1980) and t = 9 (January 1, 1989). Although we already have this information for t = 9 (see part (b)), we do not yet have this information for t = 0. Therefore, we would first need to draw a tangent line to the graph corresponding to t = 0. This is done below.

Average & Instantaneous Rates of ChangeCONTINUED Obviously, finding the coordinates of two points on this tangent line might prove a little difficult. However, notice that the slopes of the two tangent lines (all we’re really interested in are their slopes) are not remotely similar (that is, the tangent lines are not close to being parallel). Therefore, in this circumstance, it would be sufficiently appropriate to notice that the blue tangent line (corresponding to t = 0) has a steeper slope and therefore the rate of change was greater on January 1, 1980 than it was on January 1, NOTE: Use this technique of “eye-balling” a graph only when absolutely necessary and only with great care.

Average Velocity DefinitionExample Average Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is miles per minute where a = 5 and h = 6.

Position, Velocity & Acceleration s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.

Position, Velocity & Acceleration A toy rocket fired straight up into the air has height s(t) = 160t – 16t 2 feet after t seconds. EXAMPLE (a) What is the rocket’s initial velocity (when t = 0)? (b) What is the velocity after 2 seconds? (c) What is the acceleration when t = 3? (d) At what time will the rocket hit the ground? (e) At what velocity will the rocket be traveling just as it smashes into the ground?