AP Calculus

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Special Focus: The Fundamental Theorem of Calculus These values look like multiples of π ! By hand, we investigate when Si′ ( x) = 0. sin( x) Si′ ( x) = 0 ⇒ = 0 ⇒ sin( x) = 0 and x ≠ 0 ⇒ x = ± kπ for some x integer k ≠ 0 x = ± π, ± 2π, ± 3 π, ... (b) Consider these zeros and the graph of y = Si′( x) to find local maximum and local minimum values. For x >0: Local maximum values: Si( π ) =1. 8594, Si( 3π ) = 1. 6748, Si( 5π ) = 1. 6340, Si( 7π ) = 1. 6161, Si( 9π ) = 1. 6061 Local minimum values: Si( 2π ) = 1. 4182, Si( 4π ) = 1. 4922, Si( 6π ) = 1. 5180, Si( 8π ) = 1. 5311, Si( 10π ) = 1. 5390 The local maximum values are getting smaller as x increases, and the local minimum values are getting bigger as x increases. Although it’s hard to tell, even using technology, these values are converging, and the graph of y = Si( x) π has a horizontal asymptote at y = ≈ 2 1. 571. For x
Special Focus: The Fundamental Theorem of Calculus The local maximum values are decreasing as x decreases, and the local minimum values are increasing as x decreases. Again, although it’s hard to tell, even using technology, these values are converging, and the graph of y = Si( x) π has another horizontal asymptote at y = − ≈− 1. 571. 2 (c) There are a couple of ways to find this inflection point. (i) The graph of y = Si(x) changes from decreasing to increasing somewhere between x = 4 and x = 5, indicating an inflection point. Use technology to approximate this value. (ii) Find Si′′( x), set Si′′ ( x) = 0, and solve (numerically): xcos( x) − sin( x) Si′′ ( x) = = 0 . 2 x Either method produces the inflection point (4.49341, 1.65557). (d) Using technology, here is a graph of y = Si( x). 120 AP® Calculus: 2006–2007 Workshop Materials
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AP Calculus

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