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Unformatted text preview: wtm369 Homework 5 Helleloid (58250) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find the xcoordinate 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. xcoords = 2 3 , 2 correct 2. xcoord = 2 3. xcoord = 2 4. xcoord = 2 3 5. xcoord = 2 3 6. xcoords = 2 3 , 2 Explanation: The tangent line will be horizontal at P ( x , f ( x )) when f ( x ) = 0 . Now f ( x ) = 3 x 2 4 x 4 = (3 x + 2)( x 2) . Consequently, x = 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 pointslope 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 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 ....
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This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz
 Calculus

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