# HW5 - Jolley Garrett – Homework 5 – Due 3:00 am –...

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Unformatted text preview: Jolley, Garrett – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: R Heitmann 1 This print-out should have 24 questions. Multiple-choice 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 + 2 x- 6 . 1. slope = 4 h- 10 2. slope = 2 h + 6 3. slope = 4 h + 6 4. slope = 2 h- 6 correct 5. slope = 4 h + 10 6. slope = 2 h- 10 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(- 2 + h )- 6 } + 2 h = 2 h 2- 6 h h = 2 h- 6 . 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 + 4 x + 5 at which the tangent line is parallel to the line y = 3 x + 2 , determine a . 1. a =- 1 2 correct 2. a =- 1 3. a =- 5 2 4. a =- 2 5. a =- 3 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 ) = ( a + h ) 2 + 4( a + h ) + 5 = a 2 + h (2 a + 4) + h 2 + 4 a + 5 , while f ( a ) = a 2 + 4 a + 5 . Thus f ( a + h )- f ( a ) = h { (2 a + 4) + h } , in which case f ( a ) = lim h → { (2 a + 4) + h } = 2 a + 4 . If the tangent line at P is parallel to the line y = 3 x + 2 , Jolley, Garrett – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: R Heitmann 2 then they have the same slopes, so f ( a ) = 2 a + 4 = 3 . Consequently, a =- 1 2 . keywords: tangent line, parallel, slope, deriva- tive 003 (part 1 of 1) 10 points Find the x-intercept of the tangent line at the point P (2 , f (2)) on the graph of f when f is defined by f ( x ) = 3 x 2 + 2 x + 3 . 1. x-intercept =- 14 9 2. x-intercept = 9 3. x-intercept = 9 14 correct 4. x-intercept = 14 9 5. x-intercept =- 9 6. x-intercept =- 9 14 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 ) = 6 x + 2 at x = 2, i.e. , m = 14. On the other hand, f (2) = 19. Thus by the point-slope formula an equation for the tangent line at P (2 , f (2)) is y- 19 = 14( x- 2) , i . e ., y = 14 x- 9 . Consequently, x-intercept = 9 14 . keywords: tangent line, x-intercept, 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 ) = 3- 2 x 3 . 1. y + 6 x = 7 correct 2. y + 7 x + 6 = 0 3. y = 7 x- 6 4. y + 6 x + 7 = 0 5. y = 6 x + 7 Explanation: If x = 1, then g (1) = 1. Thus the Newto- nian different quotient for g ( x ) = 3- 2 x 3 at the point (1 , 1) becomes g (1 + h )- g (1) h = h 3- 2 (1 + h ) 3 i- 1 h = 3- 2 h 3- 6 h 2- 6 h- 3 h ....
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HW5 - Jolley Garrett – Homework 5 – Due 3:00 am –...

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