1110-fa2011-PRELIM-1-SOLUTIONS

# 1110-fa2011-PRELIM-1-SOLUTIONS - Math 1110 Prelim I 1...

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Math 1110 Prelim I (9/27/2011) 1 Question 1. (10 points) The National Oceanic and Atmospheric Administration collects climate data at weather stations around the US. The chart below shows the total rainfall to date (beginning June 1, 2006) observed at the Binghamton, NY, station during the month of June 2006. The dots indicate the measurement at the end of the corresponding day. 30#,4.5/6#' 78 9 0#:4 & 0#:4\$; = 0#:4\$; 9( 0#:4\$; >05\$ +56"#/ 61 -.0#1.22 (a) What was the total rainfall in Binghamton in June, 2006? The total rainfall in June, 2006, was approximately 11.3 inches. We accepted any answer that was strictly greater than 11 or less than 11.6 for full credit. (b) Estimate the average rate of rainfall from June 1 through 4, 2006. The average rate of rainfall was approximately f ( 4 ) - f ( 0 ) 4 - 0 = 2.2 4 = 0.55 inches per day . We also accepted f ( 4 )- f ( 1 ) 4 - 1 , and slightly different approximate values for f ( 4 ) . (c) On which day was the rain falling the most heavily? How can you tell? It was falling the most heavily on June 27, 2011, because the slope of the graph is the steepest on that day. (d) The graph of these data exhibits increasing behavior. Would we ever see a decrease for this function? Why or why not? You will never see a decrease because these are cumulative data: each days rainfall is added to the previous total. Since it can never “unrain,” the function either increases or remains constant. (The data do not include evaporation either!)

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Math 1110 Prelim I (9/27/2011) 2 Question 2. (28 points) Evaluate the following expressions, or explain why you cannot. You may (a) D omain ± ln ( x ) x - 1 ² = ( 0,1 ) ( 1, ) or all positive real numbers not equal to 1 . (b) cos ( log 9 ( 3 π ) ) = cos ( π log 9 ( 3 ) ) = cos ³ π · 1 2 ´ = cos ± π 2 ² = 0. (c) Find a value of x so that arcsin ( sin ( x )) 6 = x . The value x = π satisﬁes this, because arcsin ( sin ( π )) = arcsin ( 0 ) = 0 6 = π. (d) lim x - 1 x 2 - 1 | x | - 1 If x < 0 then | x | = - x and so lim x - 1 x 2 - 1 | x | - 1 = lim x - 1 x 2 - 1 - x - 1 = lim x - 1 ( x - 1 )( x + 1 ) -( x + 1 ) = lim x - 1 -( x - 1 ) = 2.
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## This note was uploaded on 03/01/2012 for the course MATH 1110 at Cornell.

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1110-fa2011-PRELIM-1-SOLUTIONS - Math 1110 Prelim I 1...

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