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140Bmid2csols

# 140Bmid2csols - 1 MATH 140B Fall 2007 Midterm 2 Solutions...

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Unformatted text preview: 1 MATH 140B Fall 2007 Midterm 2 Solutions Instructor: Dr. M.Fabbri Duration: 50 minutes 1. (This is #3 on Version B) Since the degree of the numerator is equal to the degree of the denominator, the limit is the ratio of the leading coefficients. Therefore, the answer is- 1 / 2 . Therefore, B) is the answer. 2. (This is #4 on Version B) The degree of the numerator is greater than that of the denominator, and so the value of the limit is one of . The ratio of the leading coefficients is negative, so the answer is- . Therefore, A) is the answer. 3. (This is #5 on Version B) i) True . By the 2nd derivative test. ii) True . By uniqueness of the critical point. Therefore, D) is the answer. 4. (This is #1 on Version B) i) True . Easy differentiation. ii) True . 12 x 2 = 0 implies x = 0 . iii) False . Since f ( x ) &amp;gt; 0 on either side of x = 0 , the concavity does not switch. Therefore, B) is the answer. 5. (This is #2 on Version B) x 2 yy + 1 y 2 + 3 y = 0 and so at (- 2 , 1) this becomes- 4 y + 1 + 3 y = 0 . This gives y = 1 and so the slope-point form of the tangent line is y- 1 = 1 ( x + 2). Therefore+ 2)....
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140Bmid2csols - 1 MATH 140B Fall 2007 Midterm 2 Solutions...

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