1.1 : (The Slope of a straight Line)

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1 1.1 : (The Slope of a straight Line) Equations of Nonvertical Lines: A nonvertical line L has an equation of the form y mx b. 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. y 3x 4 For this line, the slope m = 3 and the y-intercept b = -4. Example The following are graphs of equations of lines that have positive slopes. Example The following are graphs of equations of lines that have negative slopes. 1

2 Properties of the Slope of a Nonvertical Line We write: L 1 // L In another word if the slope of the first line is (m), then the slope of the second line is (- ).

3 We write: L 1 L the lines Y = x +1, Y = x, Y = x 1, are parallel lines. the line y = 3x+1 is perpendicular to the line y = - x + 1 Example Find the slope and y-intercept of the line 1 y x. 3 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. 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. Now we connect the two points that have already been determined, since two points determine a straight line. 3

4 Find an equation of the line that passes through the points (-1/, 0) and (1, ). Using the two points we will determine the slope by using Slope Property. 0 m We now use Slope Property 3 to find an equation of the line 4 Find an equation of the line that passes through the point (, 0) and is perpendicular to the line y = x. We can now find the slope of the line that we desire. Let the slope of the new line be m. Now we can find the equation of the desired line using Property 3. As we have a point (,0) and the slope m =

5 1. : (Slope of a curve at a point) Slope as a Rate of Change = = = m Compute the rate of change of the function over the given intervals. Solution : y x 7 0,1, 0,0.5 Since this is clearly a linear function by definition, it also has a constant rate of change, -. Therefore, no matter what interval is considered for this function, the rate of change will be -. Therefore the answer, for both intervals, is -. Tangent Lines Slope of a Curve & Tangent Lines 5

6 Find the slope of the tangent line to the graph of y = x write the corresponding equation of the tangent line. at the point (-0.4, 0.16) and then 6

7 1.3 : (The Derivative) Derivative: The slope formula for a function y = f (x), denoted: Differentiation: The process of computing a derivative. Differentiation Examples / y f x. Find the derivative of f 1 x. 7 x What we ve done so far has been done for the sole purpose of rewriting the function in the form of f (x) = x r. f x x x 7 x 7 Find the slope of the curve y = x 5 at x = -. We must first find the derivative of the given function 7

8 Equation of the Tangent Line to the Graph of y = f (x) at the point (a, f (a)) Find the equation of the tangent line to the graph of f (x) = 3x at x = 4. Leibniz Notation for Derivatives 8

9 1.4 : ) Limit and the Derivatives) Definition of the Derivative: And, (a) = (VII) Limit of a polynomial Function : Let P(x) be a polynomial function, a any number. Then (VIII) Limit of a Rational Function: Let r(x)= be a rational function, where p(x) and q(x) are polynomials. Let a be a number such that q(a) 0. Then Examples: (1) ( 5 x3 15) = 5() 3 15= 5 () = = (x+3) = (3+3)= 6 (3), Since the denominator approaches zero when taking the limit, we need to simplify the expression. = = = = 9

10 Thus, = = = (4) 10x 100 lim x x 30 Both10x+ 100 and x 30 increase without bound as x does. To determine the limit of their quotient, we employ an algebraic trick. Divide both numerator and denominator by x (since the highest power of x in either the numerator or the denominator is ) to obtain x 100 lim lim x x. x x 30 x 30 1 x As x increases without bound, 10/x approaches 0, 100/x approaches 0, and 30/x approaches 0. Therefore, as x increases without bound, 10/x + 100/x approaches = 0 and 1-30/x approaches 1 0 = 1. Therefore, x lim lim x x x 30 x x x 0. 11

11 Use limits to compute the derivative f / (3) for the function f x 1 x 5. 11

12 Thus, f / (-1) = -(-1)= 1

13 Example 13

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15 1.6 : ( Some Rules for Differentiation) 15

16 (1) () (3) 16

17 (4) The Derivative as a Rate of Change Let S(x) represent the total sales (in thousands of dollars) for month x in the year 005 at a certain department store. Represent each statement below by an equation involving S or S (a) The sales at the end of January reached $10,560 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 $00 a day (Use 1 month = 30 days). (a) Since the sales at the end of January (the first month, so x = 1) reached $10,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 10,560. 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: S (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 $00 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-00 dollars) = dollars per month. Therefore, we have: S

18 1.7: More About Derivatives (Second Derivatives) The second derivative, is defined to be the derivative of the derivative. We write: f // (x) = or we write, f(x) to denote the second derivative of f(x) 18