Section 2.4: Definition of Function

Transcription

1 Section.4: Definition of Function Objectives Upon completion of tis lesson, you will be able to: Given a function, find and simplify a difference quotient: f ( + ) f ( ), 0 for: o Polynomial functions of degree less tan 4, o Rational functions, and o Square root functions. Sketc te grap of a polynomial or square root function and find o Te domain and range, and o Intervals on wic te function is increasing, decreasing, or constant. Construct a model in terms of one variable and use te model to answer questions. o Linear o Volume o Area/Surface Area o Perimeter/Circumference o Distance d r t d ( 1) + ( y y1) o Pytagorean Teorem o Cost Required Reading Swokowski/Cole: Section.4, pages Please review te details of interval notation in Section 1.6, Inequalities, pages Te cart on page 76 is ecellent. Discussion Te Difference Quotient Te difference quotient is simply te slope of te secant line connecting two points on a curve. Refer to Figure 9 on te bottom of page 144 and te figure below.

2 f f(+) secant line f() + For te function f in te figure above, te slope of te secant line is y ms y y 1 1 f ( + ) f ( ) ( + ) f ( + ) f ( ), 0 Te tet uses multiple versions of te difference quotient. Tey include: f ( a + ) f ( a), 0; f ( ) f ( a), a ; and a f ( + ) f ( ), 0. Study eample 5 on page 145 of te tet. It is an ecellent eample of a polynomial function. Te net tree eamples will demonstrate tecniques to use wen simplifying a difference quotient for polynomial, rational, and square root functions. Eample 1: Find and simplify te difference quotient f ( + ) f ( ) for f ( ) + 3. Solution: f ( + ) ( + ) + 3 ( + ) f ( + ) f ( ) ( ) ( + ) + 3( + ) ( + 3) Epand and distribute. Combine like terms. Factor out te term and simplify. + 3

3 Eample : Find and simplify te difference quotient f ( + ) f ( ) 1 Solution: f ( + ) for 1 f ( ). + 3 f ( + ) f ( ) ( 3)( 3) ( + + 3)( + 3) Multiply by te common denominator. ( + 3) ( + + 3) ( + + 3)( + 3) ( + + 3)( + 3) 1 ( + + 3)( + 3) Distribute and combine like terms. Simplify. Eample 3: Find and simplify te difference quotient f ( + ) f ( ) for f ( ). Solution: f ( + ) ( + ) f ( + ) f ( ) ( + ) ( + ) ( + ) + ( + ) + Multiply by te conjugate of te numerator. ( + ) ( ( + ) + ) ( ( + ) + ) ( + ) + Distribute and combine like terms. Simplify. 3

4 Word Problems You may find te word problems in tis section very callenging, most students do. On te inside front cover of your tet and te front and back of te net page, you will find most of te formulas needed to complete tis omework. Below are four eamples of word problems to complement te tet. Eample 4: Epress te area of a circle in terms of its circumference. Solution: Te formula for te area of a circle is A A( r) πr. Te formula for te circumference of a circle is C C( r) π r. Since we want area as a function of circumference, we will solve for r in te circumference formula and substitute tat result in to te area formula. C C C C r and A A C π π ( ) π π 4π 4π Wit tis result, if we know te circumference of a circle, we can directly compute its area. Eample 5: Let P (, y) be a point on te grap of y 8. Epress te distance d from P to te point (1, 0) as a function of. Solution: Te distance between te points (, y) and (1, 0) is given by d ( 1) + ( y 0) ( 1) + y. Wen working tese problems, try te following tecniques: 1. Draw a picture to represent te problem if possible.. Write a formula to represent te unknown in terms of te available variables. 3. Substitute to get te function in terms of a single variable. Tis is a function of bot and y. Since y 8 describes te relationsip between and y,. we substitute and ave d( ) ( 1) + ( 8) Eample 6: An open bo wit a square base is required to ave a volume of 10 cubic feet. Epress te amount S of material used to make suc a bo as a function of te lengt of a side of te square base. Solution: Te volume of te bo is V or 10 or 10, were is te lengt of a side and is te eigt of te bo. Te amount of material used is te surface area and given by S + 4, were is te surface area of te base and 4 is te surface area of te four sides. S is a function of bot 4

5 10 and. We will now substitute for in terms of. Tis gives S + 4 or 40 S( ) +. Eample 7: A small island is miles from te nearest point A on te straigt soreline of a large lake. A woman on te island wises to reac a town 10 miles down te sore from point A. Te woman can row a boat 3 mp and can walk 4 mp. Te woman will row to a point P on te soreline and walk miles to te town. If T is te total time required to reac te town, epress T as a function of. Solution: Sketc te information given above. Te sketc represents te distance of eac leg of te trip from te island diagonally to te soreline and ten down te soreline to te town. Te distance from te island to te soreline is d1 + ( 10 ). Te distance down te soreline from P to te town is d. Since distance (rate)(time), time is distance/rate. Te time to make tis trip is te sum of te time it takes to row and te time it takes to walk. d1 d 4 + ( 10 ) Hence, T Practice Problems Work tese problems. Answers to te odd numbered problems can be found at te end of your tet, even answers are below. Section Pages Eercises f, 7f, 9f, 37, 39, 41, 43, 47, 49, 50, 51, 5, 65, 66, 67, 68, 69, 71, 73, 74, 75, 76, 77, 78, 86, 87 Answers to even eercises ( + ) 5. + a + a 66. S ( r) 4π r(5 + r) 68a. 68b. 4 y( ) 1 S( ) a. y( ) b. A ( ) 5 ; (only 0 < < 15 will form triangles) 5

6 76a. L ( ) ( ) 76b ft T ( ) b. No 86c. Doubling te speed requires almost five times te stopping distance. 6