The above information is from your computer : be sure it is correct. For grade credit, submit over the internet by 11:59 PM (= 23:59 hours) on Sunday 28 October 2017

Self-marking Practice quiz 4

(Relative Extrema , Anti-derivatives)
Submit over the internet by 11:59 PM Sunday 28 October 2017
Maximum value toward semester grade is 4 pts
Methods will be discussed in class;
CAUTION: Prof McFarland makes new tests each semester.
 Average grade on Fall 2017 paper test #4 : 27.83 perfect = 45 See future grade prospects?

 Click or type the appropriate answer for each question. When you are sure of your answers, send them to your Professor: Enter your name above Enter preferred email address;include @ and whatever follows Enter student I.D.
 [1] (a)(1 pt) Find f '(x) if f(x) = x3 - 3x2 - 9x + 1 . f '(x) = x3 + x2 + x + (b)(4 pts) Find both CVs (critical values) for f(x) in [1a] above. Then test each CV by entering numbers in the 12 empty boxes of the table below. Scoring: 1 point for each CV (top two answer boxes) ; 1 point for each of the remaining 2 rows below
 Note that the table at the right is a 1st derivative test
 test value CV test value CV test value x -2 +1 +4 f '(x) 0 0 Type of CV : Relative Maximum, Relative Minimum,Ledge Max Min Ledge Max Min Ledge

[2]
 Let f(x) = x3 - 3x2 - 9x + 1 with the domain of f restricted to: 0x+2 : Compare carefully with [1] above.
(a)(1 pt) Enter BOTH CV's for the above f(x) into the 2nd row of the table below, as well as a test value between.
(b)(4 pts) Fill in the remaining empty boxes in the 3rd, 4th, and 5th rows of the table below.
Thus, you will display both a graph test and a 1st derivative test for these two CV's.
Scoring : 1 point for each answer in bottom row ; ½ point each for other table entries

 smaller CV Test Value larger CV x +1 f (x) f '(x) Type of CV : Relative Maximum, Relative Minimum,Ledge Max Min Ledge Max Min Ledge

[3]
 An open-topped box is to be made from an 6 cm by 8 cm square of cardboard, by removing a smaller square from each corner, and folding up the resulting flaps, as in the animation below. What size of cut-out square will produce the box of greatest volume?
 "I found a BIG HINT on Prof's website!"

 (a)(2 points) For the problem enclosed above, let "x" be the edge length of cut-out squares, as shown, and let V be the volume of the resulting box. Enter integers below (such as 3, 0, or -2), expressing the relationship between "x" and "V". Thus, if you (incorrectly) thought V = (10 - x)(x + 2), then you would multiply out the product and enter "0", "-1", "19", and "20" in the boxes below. V (x) = x3 + x2 + x +
(b)(3 points) Put aside (for the moment) any physical restrictions on x, and find the two values of x at which ' = 0 ; you will need the quadratic formula to find these numbers. Enter them as CVs in the table below, and complete the 2nd derivative test below, thus answering the question in the problem above. Scoring : ½ point for each of 6 correct table boxes below
 Note that the table at the right is a 2nd derivative test
 smaller CV larger CV x V "(x) Type of CV : Relative Maximum, Relative Minimum,Ledge Max Min Ledge Max Min Ledge
As to those "physical restrictions" : cut yourself a 6" by 8" rectangle and use the larger CV above to build a box ; what happens?
(unscored)

 [4] (a)(3 pts) If q represents the number of lamps sold, and if p represents the price of each lamp, and if p and q are related by the equation : q = 60 - 0.1p , then calculate the elasticity of demand when p = 200. [See sample problem] elasticity of demand = (b)(2 pts)) The gross national (GNP) product of Singapore was \$100 billion in 2000, and \$200 billion in 2010. If Singapore's GNP is growing exponentially over time, what should it be in 2020? [See sample problem] Singapore's GNP = \$ billion

[5]
 (3 pts) Find f '(x) and f "(x) if :   f(x) = xx
AFTER finding these derivatives, let "x" = 1, to simplify your entries below.
(a)(3 pts) f '(1) =
(b)(2 pts) f "(1) =

[6]
 In [5b] above, you found f "(x) when f(x) = xx
Use the SCORE CHECKER below to be sure answer to problem [5b] is correct.
Use this f "(x) to find the only 2nd order CV for f(x), that is, CANDIDATE. Then test your candidate as to whether or not it is indeed an IP (inflection point) by filling in the table below (accurate to within 0.0001).
Scoring : 1 point for each table box below
 test value candidate test value x 1 5 f "(x) test results :check one IP not an IP
See a similar but somewhat harder problem in assignment #7

[7]
 (a)(1 pt) First find F(x) = [x - 2 x -3 ] dx .
Finally, let C=0 , x=1 , and enter F(1) :

 (b)(2 pts) First find F(x) = dx (integrand is multiplied by ).
Finally, let C=0 , x=1 , and enter F(1) :

 (c)(2 pts) First find F(x) = dx (integrand is divided by ).
Finally, let C=0 , x=1 , and enter F(1) :

[8]   Find   18x
 3x2 - 8
dx   =   F(x) + C ;
As you use the Substitution Method, you will alter the integrand so that u' appears as a factor. Identify your substitution variables u and y and their derivatives in the table at the right.
Finally, let C = 0, x = 2, and enter F(2) =
Scoring : 3 pts for F(2) above ; 2 pts for for table (right)
 For example, if this question were to find 8x3 dx , then F(x) = 2x4 + C , and you would enter F(2) = 32
 u =
x

x

3x2 - 8

3x2

x2
 dudx =
1
3x2 - 8

18x
6x
2x
 y =
u

u2

2( )3
 dydu =
u

x

3

 [9] Find (1 - x) e x dx   =   F(x) + C ;
As you use Integration by Parts to find F(x), you must choose expressions u, v, u', and v' from alternatives in the table at the right.
Finally, let C = 0, x = 1, and enter F(1) =
Scoring : 3 pts for F(1) above ; 2 pts for for table (right)
 For example, if this question were to find 8x3 dx , then F(x) = 2x4 + C , and you would enter F(1) = 2
 u =
1
x
1 - x
e
ex
(1 - x)ex
 u ' =

=
0
1
-1
e
ex
-xex
 v =
1
x
1 - x
e
ex
(1-x)ex
 v ' =

=
0
1
-1
e
ex
-xex

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