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Homework 5-solutions - wtm369 Homework 5 Helleloid(58250...

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wtm369 – Homework 5 – Helleloid – (58250) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the x -coordinate of all points on the graph of f ( x ) = x 3 2 x 2 4 x + 5 at which the tangent line is horizontal. 1. x -coords = 2 3 , 2 correct 2. x -coord = 2 3. x -coord = 2 4. x -coord = 2 3 5. x -coord = 2 3 6. x -coords = 2 3 , 2 Explanation: The tangent line will be horizontal at P ( x 0 , f ( x 0 )) when f ( x 0 ) = 0 . Now f ( x ) = 3 x 2 4 x 4 = (3 x + 2)( x 2) . Consequently, x 0 = 2 3 , 2 . 002 10.0 points The normal to the graph of a function f at a point P on the graph is the straight line through P perpendicular to the tangent to the graph of f at P . Find the equation of the normal to the graph of f ( x ) = 5 x 2 6 x + 8 at P (1 , f (1)). 1. y + 4 x = 11 2. y 4 x 3 = 0 3. y + 1 4 x = 15 2 4. y 1 4 x = 27 4 5. y + 1 4 x = 29 4 correct Explanation: When x = 1, we see that f (1) = 7. Thus P = (1 , 7). On the other hand, the slope of the tangent line to the graph of f at (1 , 7) is the value of the derivative f ( x ) = 10 x 6 at x = 1, i.e. , f (1) = 4. Now a line per- pendicular to a line having slope m has slope 1 /m . Thus at (1 , 7) slope of normal = 1 4 . Consequently, by the point-slope formula, the equation of the normal is given by y 7 = 1 4 ( x 1) , which in turn can be rewritten as y + 1 4 x = 29 4 . 003 10.0 points If P is a second degree polynomial such that P (2) = 8 , P (2) = 4 , P ′′ (2) = 1 , find the value of P (1). 1. P (1) = 7 2
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wtm369 – Homework 5 – Helleloid – (58250) 2 2. P (1) = 4 3. P (1) = 3 4. P (1) = 9 2 correct 5. P (1) = 5 Explanation: The most general second degree polynomial has the form P ( x ) = ax 2 + bx + c . But then P ( x ) = 2 ax + b, P ′′ ( x ) = 2 a . Thus P ′′ (2) = 1 = a = 1 2 , in which case P ( x ) = x + b . The condition on P (2) now ensures that P (2) = 2 + b = 4 , i.e. , b = 2. Thus P ( x ) = 1 2 x 2 + 2 x + c , in which case P (2) = 8 = c = 2 . Consequently, P ( x ) = 1 2 x 2 + 2 x + 2 . At x = 1, therefore, P (1) = 9 2 . 004 10.0 points Find the derivative of f ( x ) = 3 x 1 4 2 x 1 4 + 4 . 1. f ( x ) = 3 x 1 2 + 2 4 x 5 4 correct 2. f ( x ) = 3 x 1 4 + 2 4 x 3 4 3. f ( x ) = 3 x 1 2 + 2 3 x 5 4 4. f ( x ) = 3 x 1 2 2 4 x 5 4 5. f ( x ) = 3 x 1 2 2 4 x 3 4 Explanation: Since d dx ( x r ) = rx r 1 , we see that f ( x ) = 1 4 parenleftbigg 3 x 3 4 + 2 x 5 4 parenrightbigg . Consequently, f ( x ) = 3 x 1 2 + 2 4 x 5 4 . 005 10.0 points Find the derivative of g when g ( t ) = 5 4 t 4 . 1. g ( t ) = 5 t 5 correct 2. g ( t ) = 5 t 5 3. g ( t ) = 5 4 t 5 4. g ( t ) = 5 t 3
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wtm369 – Homework 5 – Helleloid – (58250) 3 5. g ( t ) = 5 4 t 5 6. g ( t ) = 5 4 t 3 Explanation: Since d dt a t r = r a t r 1 for all r negationslash
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