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Unformatted text preview: Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken 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 ( 1 , f ( 1)) , ( 1 + h, f ( 1 + h )) when f ( x ) = 2 x 2 3 x 4 . 1. slope = 2 h + 7 2. slope = 2 h 1 3. slope = 4 h + 7 4. slope = 4 h 1 5. slope = 2 h 7 correct 6. slope = 4 h + 1 Explanation: Since the secant line passes through the points ( 1 , f ( 1)) , ( 1 + h, f ( 1 + h )) , its slope is given by f ( 1 + h ) f ( 1) h = { 2( 1 + h ) 2 3( 1 + h ) 4 } 1 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 ) = 2 x 2 + 3 x + 6 at which the tangent line is parallel to the line y = 6 x + 4 , determine a . 1. a = 1 2. a = 1 4 3. a = 3 4 correct 4. a = 5 4 5. a = 1 2 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 ) = 2( a + h ) 2 + 3( a + h ) + 6 = 2 a 2 + h (4 a + 3) + 2 h 2 + 3 a + 6 , while f ( a ) = 2 a 2 + 3 a + 6 . Thus f ( a + h ) f ( a ) = h { (4 a + 3) + 2 h } , in which case f ( a ) = lim h → { (4 a + 3) + 2 h } = 4 a + 3 . If the tangent line at P is parallel to the line y = 6 x + 4 , Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken 2 then they have the same slopes, so f ( a ) = 4 a + 3 = 6 . Consequently, a = 3 4 . 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 ) = 2 x 2 3 x + 2 . 1. xintercept = 6 11 2. xintercept = 11 6 3. xintercept = 6 11 correct 4. xintercept = 11 6 5. xintercept = 6 6. xintercept = 6 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 ) = 4 x 3 at x = 2, i.e. , m = 11. On the other hand, f ( 2) = 16. Thus by the pointslope formula an equation for the tangent line at P ( 2 , f ( 2)) is y 16 = 11( x + 2) , i . e ., y = 11 x 6 . Consequently, xintercept = 6 11 . 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 x 3 . 1. y + 3 x + 7 = 0 2. y = 7 x 3 3. y + 7 x + 3 = 0 4. y = 3 x + 7 5. y + 3 x = 7 correct Explanation: If x = 1, then g (1) = 4. Thus the Newto nian different quotient for g ( x ) = 5 x 3 at the point (1 , 4) becomes g (1 + h ) g (1) h = h 5 (1 + h ) 3 i 4 h = 5 h 3 3 h 2 3 h 5 h ....
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This note was uploaded on 09/28/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
 schultz

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