c) Words: The cost of a taxicab is $2.00 for the first 1/4 of a mile and $1.00 for each additional 1/8 of a mile.

Transcription

1 Functions Definition: A function f, defined from a set A to a set B, is a rule that associates with each element of the set A one, and onl one, element of the set B. Eamples: a) Graphs: b) Tables: c) Words: The cost of a taicab is $.00 for the first / of a mile and $.00 for each additional /8 of a mile. d) Smbols: f ( ) = + MATH 380 Lecture of Ronald Brent 08 All rights reserved.

2 Definition: Eamples: The set A of all inputs is called the domain, and the set B of all outputs is called the range. Domain Range Eample: f ( ) = + Domain {0, 0, 0, 30, 0, 50, 60} Range {50, 57.5, 65, 80, 95, 5, 00} Domain All real or (, ) Range All real or [, ) MATH 380 Lecture of Ronald Brent 08 All rights reserved.

3 Graphs Definition: The Cartesian (rectangular) coordinate plane consists of ordered pairs of numbers (, ), or points, where the value of indicates the horizontal position of the point and the value refers to the vertical position. b (a, b) a Coordinate Aes -ais Points (a, 0) -ais Points (0, b) II I Quadrants: I All points (a, b) with a > 0 and b > 0. II All points (a, b) with a < 0 and b > 0. III All points (a, b) with a < 0 and b < 0. III IV IV All points (a, b) with a > 0 and b < 0. MATH 380 Lecture 3 of Ronald Brent 08 All rights reserved.

4 Eample: (, ) ( 3, ) (0, ) (3, ) (3, 0) MATH 380 Lecture of Ronald Brent 08 All rights reserved. (, )

5 Graphs of Functions Definition: The graph of a function ( or equation) = f () is the set of points (, ) that satisf the function (equation). Eample: f ( ) = f () MATH 380 Lecture 5 of Ronald Brent 08 All rights reserved.

6 Elementar Functions ) Polnomial E: ( ) = 3 f or f ( ) = + + ) Rational E: 3 + f ( ) = or f ( ) = ) Algebraic E: f ( ) = + or f ( ) = ) Eponential E: f ( ) = or f ( ) = e 5) Logarithmic E: f ( ) = ln ( ) or f ( ) = log ( ) 6) Trigonometric E: f ( ) = sin or f ( ) = cos MATH 380 Lecture 6 of Ronald Brent 08 All rights reserved.

7 Polnomial Functions Definition: An function of the form f = a + a + a + a + a + + a n 3 n ( ) 0 3 is a0, a, a, a3, a, an are called the called a polnomial in. the constants coefficients of the polnomial. The number n is called the degree of the polnomial. Eamples of Polnomials: Degree n = 0, f ( ) = a0 (Constant) n =, ( ) = a + a (Linear/lines) f 0 n =, n = 3, f ( ) + = a0 + a a (Quadratic/Parabolas) f ( ) + 3 = a0 + a + a a3 (Cubic) MATH 380 Lecture 7 of Ronald Brent 08 All rights reserved.

8 Constant Functions (Boring!) = a 0 f ( ) = a 0 Linear Functions (Crucial!) ( ) = a + a more familiarl known as f ( ) = m + b f 0 Definition: The slope m of the line containing the two points (, ) and (, ) is rise m = = run (, ) (, ) rise run MATH 380 Lecture 8 of Ronald Brent 08 All rights reserved.

9 Eample: Find the slope of the line containing (, 3) and (, 9) m = = = or m = = = 3 3. Eample: Find the slope of the line containing (, 3) and ( 7, 0). m = ( ) = 7 9 Eample: Find the slope of the line containing (, ) and ( 3, ). m = 3 = 0 7 = 0 Eample: Find the slope of the line containing ( 8, ) and ( 8, 6). 6 m = = = undefined. MATH 380 Lecture 9 of Ronald Brent 08 All rights reserved.

10 m = m = m = m = m = = m, for several m. MATH 380 Lecture 0 of Ronald Brent 08 All rights reserved.

11 Eample: Graph h ( ) =. If < 0, h ( ) =. If 0, h ( ) =. Put it together, and we get. MATH 380 Lecture of Ronald Brent 08 All rights reserved.

12 Theorem: The equation of the line of slope m going through the point (, ) = m( ) is Proof: Consider the picture. (, ) L (, ) 0 0 The point (, ) lies on the line L if, and onl if, the slope calculated from (, ) and 0 ( 0, 0 ) is correct, that is: m. In other words = m, which means 0 = m( 0 ). 0 (Notice that and are the variables, while m, 0, and 0 are just numbers.) This is called the point-slope form of the equation of a line. MATH 380 Lecture of Ronald Brent 08 All rights reserved.

13 All non-vertical lines cross the -ais. If a line crosses the -ais at the point (0, b), then b is called the intercept. To find the equation of the line of slope m and intercept b is eas. We use the point slope form of the equation of a line, where the slope is m, and the line contains the point (0,b). Doing so we get b = m( 0 ) which reduces to = m + b. This is called the slope-intercept form. You should also memorize this. Eample: Find the slope-intercept form of the equation of the line through the points (, ) and (3, 8). 8 6 First we compute the slope. m = = = 3. Now we use either of the two given 3 points in the point-slope form to get ( ) = 3( ). Solving for gives = 3. MATH 380 Lecture 3 of Ronald Brent 08 All rights reserved.

14 Eample: Find the slope-intercept form of the equation of the line through the points (, 3) and (8, 6). ( 6) 3 9 First we compute the slope. m = = = 8 ( ) 9. Now we use either of the two given points in the point-slope form to get ( 3) = ( )( + ) Solving for gives = +. Eample: Find the slope-intercept form of the equation of the line through the points (0, π) and (, π). π π π π First we compute the slope. m = =. So = + b Now we use either 0 π π of the two given points in the to get π = 0 + b = b so = + π MATH 380 Lecture of Ronald Brent 08 All rights reserved.

15 Parallel and Perpendicular Lines If line L has slope m, and line L has slope m, then, L is parallel to L if, and onl if, m = m, and L is perpendicular to L if, and onl if, m =. Consider the following. m Eample: Find the equation of the line passing through the point (, 5) that is parallel to the line 8 = 5. First, we determine the slope of the line given b putting it in slope-intercept form. We solve 8 = 5 for to get: 5 =. Comparing this to = m + b, we see that the slope of the given line is. Since the point on the line is (, 5), we use the point-slope form to get ( 5) = ( ). MATH 380 Lecture 5 of Ronald Brent 08 All rights reserved.

16 Eample: Find the equation of the line passing through the point (, 3) that is perpendicular to the line 6 + = 3. First, we determine the slope of the line given b putting it in slope-intercept form. We solve 6 + = 3 for to get: 3 3 =. Comparing this to = m + b, we see that the slope of the given line is slope of the perpendicular line is the negative reciprocal of on the line is (, 3), we use the point-slope form to get ( ( 3)) = ( ) or 3 7 = So, the 3, or. Since the point 3 MATH 380 Lecture 6 of Ronald Brent 08 All rights reserved.

17 Power Functions Part A: = r, with r =,, 6, etc. = 6 = = 0 Notice that these graphs seem smmetric about the -ais. It doesn't matter whether we insert or into an even power function; the answer is the same. (This shows smmetr about the -ais.) MATH 380 Lecture 7 of Ronald Brent 08 All rights reserved.

18 Part B: = r, with r = 3, 5, 7, etc. Smmetr through the origin is ehibited b the odd power functions. That is, if the point (, ) is on the graph, then so is (, ). = 7 = = MATH 380 Lecture 8 of Ronald Brent 08 All rights reserved.

19 Part C : For = r, with r =,,, 6 etc., and r = 3,,, etc. 5 7 = and, the graphs look like that in Figure 5. Since = n is equivalent to the nth root, if n is even must not be negative. The other graphs of = n for n even are ver similar to those shown here. If n is odd sa for eample = 3, or = 5 then can be an real number. The graphs of these functions are shown below. Defined onl for 0 = = MATH Lecture 9 of Ronald Brent 08 All rights reserved. 0

20 = 3 = 5 0 Defined for all The other graphs of situation. = n for n odd are ver similar to those shown here, with the same ordering MATH 380 Lecture 0 of Ronald Brent 08 All rights reserved.

21 Part D: = r, with r =, 3, etc., and,, etc. What if r is a negative integer? Can ou guess that the situation will divide up into two cases, r even or odd? Here s the situation for r odd:,, 3 etc. = 3 = For r = and, the graph is shown below. MATH 380 Lecture of Ronald Brent 08 All rights reserved.

22 = 3 = MATH 380 Lecture of Ronald Brent 08 All rights reserved.