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1.03.GrowthRates - @x x 0.5D is Its average growth rate in...

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Growth Authors: Bill Davis, Horacio Porta and Jerry Uhl ©1996-2007 Publisher: Math Everywhere, Inc. Version 6.0 1.03 Instantaneous Growth Rates BASICS B.1) Instantaneous growth rates Here is a friendly function f @ x D = 1 + 2x 3 - x 4 and a plot: Clear @ f, x D ; f @ x _ D = 1 + 2x 3 - x 4 ; fplot = Plot @ f @ x D , 8 x, - 1, 2 < , PlotStyle Ø 88 Thickness @ 0.01 D , Blue << , AspectRatio -> 1 ê GoldenRatio, AxesLabel Ø 8 "x" , "" <D - 1.0 - 0.5 0.5 1.0 1.5 2.0 x - 2 - 1 1 2 · B.1.a.i) Measure the net growth of f @ x D = 1 + 2x 3 - x 4 over the interval @ - 1,2 D . Then measure the average growth rate of f @ x D over the interval @ - 1,2 D . · Answer: Here you go: Over the interval @ - 1,2 D , the function starts out at: f @ - 1 D - 2 And it ends up at: f @ 2 D 1 Its net growth is: fgrowth = f @ 2 D - f @ - 1 D 3 Its average growth rate in units on the y -axis per unit on the x -axis over the interval @ - 1,2 D is: xgrowth = 2 - H - 1 L 3 fgrowth xgrowth 1 As x grows from - 1 to 2 , on the average, f @ x D grows at a rate of 1 unit every time x grows by 1 unit. The average growth rate of f @ x D over the interval @ - 1,2 D is 1 . · B.1.a.ii) Measure the average growth rate of f @ x D = 1 + 2x 3 - x 4 over the interval @ x,x + 0.5 D . Interpret the result. · Answer: Clear @ f, x D ; f @ x _ D = 1 + 2x 3 - x 4 1 + 2x 3 - x 4 Over the interval @ x,x + 0.5 D , the function starts out at: f @ x D 1 + 2x 3 - x 4 And it ends up at: f @ x + 0.5 D 1 + 2 H 0.5 + x L 3 - H 0.5 + x L 4 Its net growth over the interval @ x,x + 0.5 D is: fgrowth = Expand @ f @ x + 0.5 D - f @ x DD 0.1875 + 1.x + 1.5x 2 - 2.x 3 Its average growth rate in units on the y -axis per unit on the x -axis over the interval @ x,x + 0.5 D is: xgrowth = 0.5; Clear @ fAverageGrowthRate, x D ; fAverageGrowthRate @ x _ D = Expand B fgrowth xgrowth F 0.375 + 2.x + 3.x 2 - 4.x 3 On the interval @ x,x + 0.5 D , the average growth rate of f @ x D is 0.375 + 2x + 3x 2 - 4x 3 . For instance, when you look at: fAverageGrowthRate @ 0 D 0.375 then you see that, on the average,, f @ x D grows 0.375 times as fast as x grows as x advances from 0 to 0 + 0.5 = 0.5 . But when you look at: fAverageGrowthRate @ 1.5 D - 3.375 Then you see that, on the average, f @ x D goes down 3.375 times as fast as x grows as x advances from 1.5 to 1.5 + 0.5 = 2 . · B.1.a.iii) Given a positive number h , measure the average growth rate of f @ x D = 1 + 2x 3 - x 4 over the interval @ x,x + h D . Interpret the result. · Answer: Clear @ f, x D ; f @ x _ D = 1 + 2x 3 - x 4 1 + 2x 3 - x 4 Over the interval @ x,x + h D , the function starts out at: f @ x D 1 + 2x 3 - x 4 And it ends up at: Clear @ h D ; Expand @ f @ x + h DD 1 + 2h 3 - h 4 + 6h 2 x - 4h 3 x + 6hx 2 - 6h 2 x 2 + 2x 3 - 4hx 3 - x 4 Its net growth over the interval @ x,x + h D is: fgrowth = Expand @ f @ x + h D - f @ x DD 2h 3 - h 4 + 6h 2 x - 4h 3 x + 6hx 2 - 6h 2 x 2 - 4hx 3 Its average growth rate in units on the y -axis per unit on the x -axis over the interval @ x,x + h D is: xgrowth = h; Clear @ fAverageGrowthRate, x, h D ; fAverageGrowthRate @ x _ , h _ D = Expand B f @ x + h D - f @ x D h F 2h 2 - h 3 + 6hx - 4h 2 x + 6x 2 - 6hx 2 - 4x 3 This gives you the measurement of the average growth rate of f @ x D on the interval @ x,x + h D .
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