Exam2.solutions

# Exam2.solutions - Math 135 Section F1 Exam 2 Solutions 1...

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Math 135, Section F1, Exam 2 Solutions 1 . Consider the equation: x 2 y 3 + 3 = 5y 2 + x a . Find ) ( x y from this equation.   Implicitly differentiating, we obtain: 1 10 ) 2 ( ) 3 ( 3 2 2 + = + y y x y y y x ;   3 2 2 2 1 ) 10 3 ( xy y y x y - = -   y y x xy y 10 3 2 1 2 2 3 - - = b.  Find an equation of the tangent line to the graph at (2, 1). Any form of the equation is acceptable. (remember: the “y = mx + b” form is usually harder to find) 2 3 10 ) 1 )( 4 ( 3 ) 1 )( 2 ( 2 1 ) 1 , 2 ( - = - - = y tangent line equation is: ) 2 ( 2 3 1 - - = - x y 2a Find the following limit.   If you attempt to apply L’Hopital’s rule, be sure to justify its use (i.e. explain why it applies). cos(0) = 1 we have a “0/0” situation, and can attempt the limit by L’Hopital’s rule: ) 7 sin( 7 ) 2 sin( 2 ) 7 cos( 1 ) 2 cos( 1 lim 0 x x x x x = - - - still “0/0” so try L’Hopital’s rule again: = 49 4 ) 0 cos( 49 ) 0 cos( 4 ) 7 cos( 49 ) 2 cos( 4 = = x x .

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2b . Suppose F(x) = [f(g(x))] 2 where f (x) and g(x) are differentiable functions and: , 5 ) 2 ( , 12 ) 5 ( , 4 ) 2 ( = - = = g f f 2 ) 2 ( , 1 ) 5 ( , 3 ) 2 ( - = - = = g f f . Decide if each of the following are true or false (T,F): i) F(5) = [f(5)] 2 = 144 __F___ F(5) = [f(g(5))] 2 since g(5) not given the statement is not necessarily true. ii) 16 ) 2 )( 1 )( 4 ( 2 ) 2 ( ) 5 ( ) 2 ( 2 ) 2 ( = - - = = g f f F ___F____ iii) 48 ) 2 )( 1 )( 12 ( 2 ) 2 ( ) 5 ( ) 5 ( 2 ) 2 ( - = - - - = = g f f F ___T____ iv) ) 2 ( F can’t be calculated with the given information ____F___ 3. Let f(x) = 2 7 9 x + . a. a.   Compute ) 1 ( f 2 2 / 1 2 7 9 7 ) 14 ( ) 7 9 ( 2 1 ) ( x x x x x f + = + = - by the chain rule. 4 7 ) 1 ( 7 9 ) 1 ( 7 ) 1 ( 2 = + = f b . Use your answer to part a . and linearization to estimate f (1.04) . The linear estimator L(x) is just the tangent line function at x = 1, i.e. the equation of the tangent line to the graph of y= f(x) at the point (1,f(1)). f(1.04) will then be approximated by the y-value on this tangent line corresponding to x = 1.04. L(x) = tangent line equation to graph of f(x) at x = 1: 2
07 . 4 ) 04 (. 4 7 4 ) 1 04 . 1 ( 4 7 4 ) 04 . 1 ( ) 04 . 1 ( ) 1 ( 4 7 4 ) 1 )( 1 ( ) 1 ( ) ( = + = - + = - + = - + = L f x x f f x L c. For the function f(x) above, it turns out that 64 63 ) 1 ( = f . Is the estimate you found in part b. likely to be an overestimate or an underestimate of f(1.04)? Give reasoning using calculus to support your answer (not just your best estimate of f(1.03) from plugging into the function.) Since the second derivative is positive for x = 1, the graph is concave up in the vicinity of (1,f(1)). So the tangent line will lie UNDER the graph, and therefore using a tangent line y-value to estimate the true value of

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## This note was uploaded on 08/10/2008 for the course MATH 135 taught by Professor Noone during the Spring '08 term at Rutgers.

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Exam2.solutions - Math 135 Section F1 Exam 2 Solutions 1...

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