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Unformatted text preview: Tran, Tony – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Samuels 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the slope of the secant line passing through the points ( 2 , f ( 2)) , ( 2 + h, f ( 2 + h )) when f ( x ) = 2 x 2 + x + 3 . 1. slope = 4 h 9 2. slope = 4 h + 7 3. slope = 4 h + 9 4. slope = 2 h + 7 5. slope = 2 h 7 correct 6. slope = 2 h 9 Explanation: Since the secant line passes through the points ( 2 , f ( 2)) , ( 2 + h, f ( 2 + h )) , its slope is given by f ( 2 + h ) f ( 2) h = { 2( 2 + h ) 2 + ( 2 + h ) + 3 } 9 h = 2 h 2 7 h h = 2 h 7 . keywords: slope, secant line 002 (part 1 of 1) 10 points If P ( a, f ( a )) is the point on the graph of f ( x ) = x 2 + 3 x + 7 at which the tangent line is parallel to the line y = 6 x + 6 , determine a . 1. a = 0 2. a = 1 2 3. a = 1 2 4. a = 3 2 correct 5. a = 1 Explanation: The slope of the tangent line at the point P ( a, f ( a )) on the graph of f is the value f ( a ) = lim h → f ( a + h ) f ( a ) h of the derivative of f at x = a . To compute the value of f ( a ), note that f ( a + h ) = ( a + h ) 2 + 3( a + h ) + 7 = a 2 + h (2 a + 3) + h 2 + 3 a + 7 , while f ( a ) = a 2 + 3 a + 7 . Thus f ( a + h ) f ( a ) = h { (2 a + 3) + h } , in which case f ( a ) = lim h → { (2 a + 3) + h } = 2 a + 3 . If the tangent line at P is parallel to the line y = 6 x + 6 , Tran, Tony – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Samuels 2 then they have the same slopes, so f ( a ) = 2 a + 3 = 6 . Consequently, a = 3 2 . keywords: tangent line, parallel, slope, deriva tive 003 (part 1 of 1) 10 points Find the xintercept of the tangent line at the point P (2 , f (2)) on the graph of f when f is defined by f ( x ) = x 2 + 2 x + 1 . 1. xintercept = 2 2. xintercept = 3 3. xintercept = 1 2 correct 4. xintercept = 2 5. xintercept = 1 2 6. xintercept = 3 Explanation: The slope, m , of the tangent line at the point P (2 , f (2)) on the graph of f is the value of the derivative f ( x ) = 2 x + 2 at x = 2, i.e. , m = 6. On the other hand, f (2) = 9. Thus by the pointslope formula an equation for the tangent line at P (2 , f (2)) is y 9 = 6( x 2) , i . e ., y = 6 x 3 . Consequently, xintercept = 1 2 . keywords: tangent line, xintercept, slope 004 (part 1 of 1) 10 points Find an equation for the tangent line to the graph of g at the point P (1 , g (1)) when g ( x ) = 5 4 x 3 . 1. y + 13 x + 12 = 0 2. y = 13 x 12 3. y + 12 x + 13 = 0 4. y = 12 x + 13 5. y + 12 x = 13 correct Explanation: If x = 1, then g (1) = 1. Thus the Newto nian different quotient for g ( x ) = 5 4 x 3 at the point (1 , 1) becomes g (1 + h ) g (1) h = h 5 4 (1 + h ) 3 i 1 h = 5 4 h 3 12 h 2 12 h 5 h ....
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This note was uploaded on 04/27/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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