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### Exercises8_3_27

Course: MATH 166, Fall 2008
School: Iowa State
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Word Count: 347

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Exercises 1 8.3 27. y = f ( x) = f ( x) = = -2 ( 2 1 . x + 2x + 5 x 2 + 2 x + 5 0 - 1 ( 2 x + 2) 2 . 2 + 2x + 5 f ( x) = 0 for x = -1. f ( x) &gt; 0 for x &lt; -1, f ( x) &lt; 0 for x &gt; -1. 1 Maximum value of f(x) is f(-1) = . 4 (x x +1 (x ) 2 + 2x + 5 ) 2 . ) (x f ( x) = -2 (x (x (x 2 2 2 + 2 x + 5 1 - ( x + 1) 2( x 2 + 2 x + 5)(2 x + 2) ) 2 = -2 = -2 = -2 = 2 + 2 x...

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Exercises 1 8.3 27. y = f ( x) = f ( x) = = -2 ( 2 1 . x + 2x + 5 x 2 + 2 x + 5 0 - 1 ( 2 x + 2) 2 . 2 + 2x + 5 f ( x) = 0 for x = -1. f ( x) > 0 for x < -1, f ( x) < 0 for x > -1. 1 Maximum value of f(x) is f(-1) = . 4 (x x +1 (x ) 2 + 2x + 5 ) 2 . ) (x f ( x) = -2 (x (x (x 2 2 2 + 2 x + 5 1 - ( x + 1) 2( x 2 + 2 x + 5)(2 x + 2) ) 2 = -2 = -2 = -2 = 2 + 2 x + 5 - 4( x + 1) 2 (x ) ) (x 2 + 2x + 5 ) 4 2 2 + 2x + 5 2 ) 3 2 + 2 x + 5 - 4( x 2 + 2 x + 1) (x + 2x + 5 ) 3 - 3x 2 - 6 x + 1 2 + 2x + 5 ) 3 + 2x + 5 f ( x) = 0 for : (x 3x 2 + 6 x - 1 ) 3 3 x 2 + 6 x - 1 = 0. x= x x x x - 6 6 2 - 4 3 (-1) . 23 - 6 48 = . 6 -64 3 = . 6 2 = -1 . 3 = .1547 or x = -2.1547. 2 = -2.1547. 3 f ( x) > 0 for x < -1 - 2 2 2 < x < -1 + -2.1547 < x < .1547. 3 3 2 f ( x) > 0 for x > -1 + = .1547. 3 f ( x) < 0 for - 1 Here R is the region of the (x, y) - plane bounded by the graph of y = f(x) = 1 , x + 2x + 5 2 the x - axis, the y - axis, and the line x = 1. Let S be the solid of revolution generated by revolving R about the x - axis. We are to find the volume of S. On the interval 0 x 1 : y = f(x) > 0 f (x) < 0, so f(x) is decreasing. 1 1 f(0) = and f(1) = . 5 8 For 0 < x < .1547, f (x) < 0, so the graph is concave downward. For .1547, (x) f > 0, so the graph is concave upward. The point (.1547, f(.1547)) is an inflection point. Solve the equation 1 x + 2x + 5 for x in terms of y. 1 = x 2 + 2 x + 5. y y= 2 1 x 2 + 2 x + 5 - = 0. y 1 - 2 22 - 4 5 - y x= . 2 1 1 - 2 4 - 16 y x= . 2 1 x = -1 - 4. y 3 Take x = -1 + 1 - 4. y 2 The volume of S is given by [ f(x)] dx 1 0 = = 1 0 1 (x 1 2 0 Let x + 1 = 2tan ( ). Then ( ( x + 1) + 2x + 5 1 2 ) 2 dx dx. + 22 ) 2 ( x + 1) 2 + 2 2 = 4 tan 2 ( ) + 4 = 4 ( tan 2 ( ) + 1) = 4 sec 2 ( ). Thus Also dx = 2sec 2 ( ) d . Hence (( x + 1) 2 + 22 ) 2 = 16 sec 4 ( ). (( x + 1) = 1 2 + 22 ) 2 dx 1 2sec 2 ( ) d 4 16 sec ( ) 1 = cos 2 ( ) d 8 1 1 + cos( 2 ) = d 8 2 1 1 = + sin ( 2 ) 16 2 1 = ( + sin ( ) cos( ) ). 16 x 2 + 2x + 5 2 x+...

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Iowa State - MATH - 166
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Iowa State - MATH - 166
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Iowa State - MATH - 166
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Exercises 10.2 5. k + 2.k =1k -51 k -5 k -5 k = k+2 k+2 1 k 5 1- k 1 0 as k . = 2 1+ k Diverges. 7. k - k - 1 . k =2 n1 11 1 k - k - 1 . k =21 1 1 1 1 1 1 1 1 1 = - + - + - + - + - 2 1 3 2 4 3 n
Iowa State - MATH - 166
Exercises 6.4 33. : x = a ( t sin(t ) ) , y = a (1 cos(t ) ) for 0 t 2. is one arch of a cycloid. (See Problem 18 and Figure 20.) Revolve about the x - axis. Area of the surface of revolution generated = 2 2 0 2y(t) [ x (t)] 2 + [ y (
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Mathematics 166 Practice Test 3 1. Determine whether or not each of the following infinite series converges. Give a reason for your conclusion in each case. 1 (a) 4 n =1 n 3 1 (b) 4 n =1 n 3 + 1 n (c) 4 n =1 n 3 + 1(d) (e) (f)e (n =1
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1 Solution Final Exam Spring 2006 1.0e -4 x dx= lim e -4 x dxb 0b 1 = lim - e - 4 x b 4 0 1 1 = lim - e - 4b + . b 4 4 1 = . 4 1 1 1 2 3 = 2 1 = 2 2 = 3. 2. n =0 1- 3 3 3. (a) nbnn =11321 x3 21
Iowa State - MATH - 166
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{VERSION 2 3 &quot;SUN SPARC SOLARIS&quot; &quot;2.3&quot; } {USTYLETAB {CSTYLE &quot;Maple Input&quot; -1 0 &quot;Courier&quot; 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE &quot;2D Math&quot; -1 2 &quot;Times&quot; 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE &quot;2D Output&quot; 2 20 &quot; 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTY
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