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Sat Sep 8 05:17:01 2007 UTC (5 years, 9 months ago) by sh002i
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```Added tags for Rogawski's "Calculus: Early Transcendentals".
```

```    1 ##DESCRIPTION
2 ## Determine concavity
3 ## ENDDESCRIPTION
4 ##KEYWORDS('derivatives', 'concave upward')
5
6 ## Shotwell cleaned
7 ## lcao , PAID on 11-24-2003
8
9 ## DBsubject('Calculus')
10 ## DBchapter('Applications of Differentiation')
11 ## DBsection('How Derivatives Affect the Shape of a Graph')
12 ## Date('6/3/2002')
13 ## Author('')
14 ## Institution('')
15 ## TitleText1('Calculus: Early Transcendentals')
16 ## EditionText1('6')
17 ## AuthorText1('Stewart')
18 ## Section1('4.3')
19 ## Problem1('64')
20 ## TitleText2('Calculus: Early Transcendentals')
21 ## EditionText2('1')
22 ## AuthorText2('Rogawski')
23 ## Section2('4.4')
24 ## Problem2('61')
25
26
27 DOCUMENT();        # This should be the first executable line in the problem.
28
30 "PGbasicmacros.pl",
32 "PGauxiliaryFunctions.pl"
33 );
34
35 TEXT(beginproblem());
37
38 BEGIN_TEXT
39 Suppose that on the interval \(I\), \(f(x)\) is positive and concave up. Furthermore, assume that \(f''(x)\) exists and let \(g(x)=(f(x))^2\). Use this information to answer the following questions. \$BR\$BR
40
42
43  \$BCENTER \$BITALIC CU \$EITALIC (concave up), \$BITALIC CD \$EITALIC (concave down), \$BITALIC f(x) \$EITALIC , \$BITALIC f'(x) \$EITALIC , \$BITALIC f''(x) \$EITALIC, \$BITALIC 0 \$EITALIC, or \$BITALIC 1 \$EITALIC. \$ECENTER
44 \$BR\$BR
45
46 \$BBOLD a.) \$EBOLD \(f''(x) > \) \{ans_rule(10) \} on \(I\). \$BR\$BR
47 \$BBOLD b.) \$EBOLD  \(g''(x)=2(A^2+B\,f''(x)) \), where
48  \(A=\)  \{ans_rule(10) \} and \(B=\)  \{ans_rule(10) \}. \$BR\$BR
49 \$BBOLD c.) \$EBOLD \(g''(x) > \)  \{ans_rule(10) \} on \(I\). \$BR\$BR
50 \$BBOLD d.) \$EBOLD \(g(x) \) is  \{ans_rule(10) \} on \(I\).
51
52 END_TEXT
53
54 \$ans1 = 0;
55 \$ans2 = "f'(x)";
56 \$ans3 = "f(x)";
57 \$ans4 = 0;
58 \$ans5 = "CU";
59 ANS(num_cmp(\$ans1));
60 ANS(str_cmp(\$ans2));
61 ANS(str_cmp(\$ans3));
62 ANS(num_cmp(\$ans4));
63 ANS(str_cmp(\$ans5));
64 ENDDOCUMENT();        # This should be the last executable line in the problem.
```