By James Stewart

Entire suggestions handbook for: unmarried VARIABLE CALCULUS Early Transcendentals seventh variation by way of Stewart.

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2{ | = 23{ | = 323{ c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, ° licated, or posted to a publicly accessible website, in whole or in part. par © Cengage Learning. All Rights Reserved. ¤ 47 48 ¤ NOT FOR SALE CHAPTER 1 FUNCTIONS AND MODELS 14. We start with the graph of | = h{ (Figure 12) and reÀect the portion of the graph in the ¿rst quadrant about the |-axis to obtain the graph of | = h|{| . 15. We start with the graph of | = h{ (Figure 13) and reÀect about the |-axis to get the graph of | = h3{ .

Since the points (32> 2) and (1> 32=5) are on the parabola, we’ll substitute 32 for { and 2 for | as well as 1 for { and 32=5 for | to obtain two equations with the unknowns d and e. (32> 2): 2 = 4d 3 2e + 1 i 4d 3 2e = 1 (1) (1> 32=5): 32=5 = d + e + 1 i d + e = 33=5 (2) 2 · (2) + (1) gives us 6d = 36 i d = 31. From (2), 31 + e = 33=5 i e = 32=5, so an equation is j({) = 3{2 3 2=5{ + 1. 9. Since i (31) = i (0) = i(2) = 0, i has zeros of 31, 0, and 2, so an equation for i is i ({) = d[{ 3 (31)]({ 3 0)({ 3 2), or i ({) = d{({ + 1)({ 3 2).

The second ¿gure shows a maximum near { = 10. (b) You need more than one window because no single window can show what the function looks like globally and the detail of the function near { = 10. I 16. The function i ({) = {2 30 3 { has domain (3"> 30]. Its graph is very steep near { = 30, so part of the graph may appear to be missing. c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, ° licated, or posted to a publicly accessible website, in whole or in part.

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