Hat04 0205
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Hat04 0205 Hat04 0205 Presentation Transcript

  • Copyright © 2007 Pearson Education, Inc. Slide 2-1 2.5 Piecewise-Defined Functions • The absolute value function is a simple example of a function defined by different rules over different subsets of its domain. Such a function is called a piecewise-defined function. – Domain of is with one rule on and the other rule on • Example Find each function value given the piecewise-defined function Solution (a) (b) (c) xxf =)( ),,( ∞−∞ xxf −=)( ),0,(−∞ xxf =)( ).,0[ ∞     > ≤+ = .0if 2 1 0if2 )( 2 xx xx xf .ofgraphSketch the(d))3((c))0((b))3((a) ffff − .123)3(thus,2)(ruletheuse,03Since −=+−=−+=≤− fxxf .220)0(thus,2)(ruletheuse,00Since =+=+=≤ fxxf .5.4)9()3()3(thus,)(ruletheuse,03Since 2 1 2 1 2 1 22 ====> fxxf .ofgraphon theare)5.4,3(and(0,2),),1,3(pointsThe f−−:Note
  • Copyright © 2007 Pearson Education, Inc. Slide 2-2 2.5 The Graph of a Piecewise-Defined Function (d) The graph of Graph the ray choosing x so that with a solid endpoint (filled in circle) at (0,2). The ray has slope 1 and y-intercept 2. Then, graph for This graph will be half of a parabola with an open endpoint (open circle) at (0,0). time.aatpieceonedrawnis 0if 2 1 0if2 )( 2     > ≤+ = xx xx xf ,2+= xy ,0≤x 2 2 1 xy = .0>x Figure 51 pg 2-117
  • Copyright © 2007 Pearson Education, Inc. Slide 2-3 2.5 Graphing a Piecewise-Defined Function with a Graphing Calculator • Use the test feature – Returns 1 if true, 0 if false when plotting the value of x • In general, it is best to graph piecewise-defined functions in dot mode, especially when the graph exhibits discontinuities. Otherwise, the calculator may attempt to connect portions of the graph that are actually separate from one another.
  • Copyright © 2007 Pearson Education, Inc. Slide 2-4 2.5 Graphing a Piecewise-Defined Function Sketch the graph of Solution For graph the part of the line to the left of, and including, the point (2,3). For graph the part of the line to the right of the point (2,3).    >+− ≤+ = .2if72 2if1 )( xx xx xf 1+= xy 72 +−= xy ,2≤x ,2>x
  • Copyright © 2007 Pearson Education, Inc. Slide 2-5 2.5 The Greatest Integer (Step) Function Example Evaluate for (a) –5, (b) 2.46, and (c) –6.5 Solution (a) (b) (c) Using the Graphing Calculator The command “int” is used by many graphing calculators for the greatest integer function.    = integerannotisifthanlessintegergreatestthe integeranisif xx xx 5− 2 7− § ¨( )f x x= § ¨x
  • Copyright © 2007 Pearson Education, Inc. Slide 2-6 2.5 The Graph of the Greatest Integer Function – Domain: – Range: • If using a graphing calculator, put the calculator in dot mode. [ ]||)( xxf = ),( ∞−∞ },3,2,1,0,1,2,3,{}integeranis{  −−−=xx Figure 58 pg 2-124
  • Copyright © 2007 Pearson Education, Inc. Slide 2-7 2.5 Graphing a Step Function • Graph the function defined by Give the domain and range. Solution Try some values of x. .1 2 1     += xy x -3 -2 -1 0 .5 1 2 3 4 y -1 0 0 1 1 1 2 2 3 }.,2,1,0,1,{israngetheand),(isdomainTheon.so and,2,42For.1then,20ifthatNotice  −∞−∞ =<≤=<≤ yxyx
  • Copyright © 2007 Pearson Education, Inc. Slide 2-8 2.5 Application of a Piecewise-Defined Function Downtown Parking charges a $5 base fee for parking through 1 hour, and $1 for each additional hour or fraction thereof. The maximum fee for 24 hours is $15. Sketch a graph of the function that describes this pricing scheme. Solution Sample of ordered pairs (hours,price): (.25,5), (.75,5), (1,5), (1.5,6), (1.75,6). During the 1st hour: price = $5 During the 2nd hour: price = $6 During the 3rd hour: price = $7 During the 11th hour: price = $15 It remains at $15 for the rest of the 24-hour period. Plot the graph on the interval (0,24]. Figure 62 pg 2-127 
  • Copyright © 2007 Pearson Education, Inc. Slide 2-9 2.5 Using a Piecewise-Defined Function to Analyze Data Due to acid rain, The percentage of lakes in Scandinavia that lost their population of brown trout increased dramatically between 1940 and 1975. Based on a sample of 2850 lakes, this percentage can be approximated by the piecewise-defined function f . (a) Graph f . (b) Determine the percentage of lakes that had lost brown trout by 1972.       ≤≤+− <≤+− = 19759601if18)1960( 15 32 19609401if7)1940( 20 11 )( xx xx xf
  • Copyright © 2007 Pearson Education, Inc. Slide 2-10 2.5 Using a Piecewise-Defined Function to Analyze Data Solution (a) Analytic Solution: Plot the two endpoints and draw the line segment of each rule. Note: Even though there is an open circle at the point (1960,18) from the first rule, the second rule closes it. Therefore, the point (1960,18) is closed. (1960,18)pointtheusgives,18,1960 (1940,7)pointtheusgives,7,1940 == == yx yx:7)1940( 20 11 +−= xy (1975,50)pointtheusgives,50,1975 (1960,18)pointtheusgives,18,1960 == == yx yx:18)1960( 15 32 +−= xy Figure 63 pg 2-128
  • Copyright © 2007 Pearson Education, Inc. Slide 2-11 2.5 Using a Piecewise-Defined Function to Analyze Data Graphing Calculator Solution (b) Use the second rule with x = 1972. (percent)6.4318)19601972( 15 32 )1972( =+−=f By 1972, about 44% of the lakes had lost their population of brown trout.