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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 Chapter 1 Functions.

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Presentation on theme: "Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 Chapter 1 Functions."— Presentation transcript:

1 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 Chapter 1 Functions

2 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 115  The Slope of a Straight Line  The Slope of a Curve at a Point  The Derivative  Limits and the Derivative  Differentiability and Continuity  Some Rules for Differentiation  More About Derivatives  The Derivative as a Rate of Change Chapter Outline

3 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 115 § 1.1 The Slope of a Straight Line

4 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 115  Nonvertical Lines  Positive and Negative Slopes of Lines  Interpretation of a Graph  Properties of the Slope of a Nonvertical Line  Finding the Slope and y-Intercept of a Line  Sketching the Graph of a Line  Making Equations of Lines  Slope as a Rate of Change Section Outline

5 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 115 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

6 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 115 Lines – Positive SlopeEXAMPLE The following are graphs of equations of lines that have positive slopes.

7 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 115 Lines – Negative SlopeEXAMPLE The following are graphs of equations of lines that have negative slopes.

8 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 115 Interpretation of a GraphEXAMPLE SOLUTION 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. First, let’s graph the line to help us understand the exercise. (80, 460)

9 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 115 Interpretation of a Graph The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she makes, her pay increases by 5 dollars. The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay. CONTINUED

10 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 115 Properties of the Slope of a Nonvertical Line

11 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 115 Properties of the Slope of a Line

12 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 115 Finding Slope and y -intercept of a LineEXAMPLE SOLUTION Find the slope and y-intercept of the line First, we write the equation in slope-intercept form. This is the given equation. Divide both terms of the numerator of the right side by 3. Rewrite Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other 1/3 is the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept. Incidentally, it was a complete coincidence that the slope and y-intercept were the same number. This does not normally occur.

13 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 115 Sketching Graphs of LinesEXAMPLE SOLUTION Sketch the graph of the line passing through (-1, 1) with slope ½. We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one unit and to the right two units to find another point on the line. (-1, 1)

14 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 115 Sketching Graphs of Lines Now we connect the two points that have already been determined, since two points determine a straight line. CONTINUED

15 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 115 Making Equations of LinesEXAMPLE SOLUTION Find an equation of the line that passes through the points (-1/2, 0) and (1, 2). To find an equation of the line that passes through those two points, we need a point (we already have two) and a slope. We do not yet have the slope so we must find it. Using the two points we will determine the slope by using Slope Property 2. We now use Slope Property 3 to find an equation of the line. To use this property we need the slope and a point. We can use either of the two points that were initially provided. We’ll use the second (the first would work just as well).

16 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 115 Making Equations of LinesCONTINUED This is the equation from Property 3. (x 1, y 1 ) = (1, 2) and m = 4/3. Distribute. Add 2 to both sides of the equation. NOTE: Technically, we could have stopped when the equation looked like since it is an equation that represents the line and is equivalent to our final equation.

17 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 115 Making Equations of LinesEXAMPLE SOLUTION Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x. To find an equation of the line, we need a point (we already have one) and a slope. We do not yet have the slope so we must find it. We know that the line we desire is perpendicular to the line y = 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the slope of the new line be m. This is Property 5. (slope of a line)(slope of a new line) = -1 The slope of one line is 2 and the slope of the desired line is denoted by m. 2m = -1 Divide.m = -0.5 Now we can find the equation of the desired line using Property 3.

18 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 115 Making Equations of LinesCONTINUED This is the equation from Property 3. (x 1, y 1 ) = (2, 0) and m = Distribute.

19 Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 115 Slope as a Rate of ChangeEXAMPLE SOLUTION Compute the rate of change of the function over the given intervals. We first get y by itself in the given equation. This is the given equation. Subtract 2x from both sides. Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope, namely -2. Therefore, by definition, it also has a constant rate of change, -2. Therefore, no matter what interval is considered for this function, the rate of change will be -2. Therefore the answer, for both intervals, is -2.


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