soltest3math1501

soltest3math1501 - Georgia Tech FALL 2009 Heinrich...

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Georgia Tech FALL 2009 Heinrich Matzinger SOLUTION FOR QUIZ 3 FOR MATH 1501: CALCULUS I 1) Let f ( x ) = x 3 3 - 3 . 5 x 2 + 6 x + 1 . Find the all the places where f ( x ) = 0. Answer: We have f ( x ) = x 2 - 7 x + 6 = ( x - 6)( x - 1). So f ( x ) = 0 leads to the two solutions x 1 = 6 and x 2 = 1. 2) Same f ( x ) as in previous problem. Where is f increasing, where is it decreasing? Answer: The term x 2 for x big in absolute value becomes dominant in comparison to 7 x . But x 2 is always positive. Hence when | x | is big enough, then f ( x ) is positive and hence the function is increasing. This implies that f ( x ) is increasing everywhere outside the interval [1 , 6]. In that interval it is decreasing. 3) Same f ( x ) as in problem 1. Where is f ( x ) concave down and where is it concave up? Answer: We have to consider the second derivative f ′′ ( x ) = 2 x - 7 We ±nd that f ′′ ( x ) is 0 at x = 3 . 5. Below 3 . 5 it is negative and hence f ( x ) is concave down there. Above 3
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soltest3math1501 - Georgia Tech FALL 2009 Heinrich...

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