MATH 151 Midterm: MATH 151 TAMU Y2013 2013a Exam 1b Solutions

• School: Texas A&M University
• Department: Mathematics
• Course: MATH 151
• Professor: All

Document Summary

Exam i version b solutions: d let f (x) = 4x2 + 16x 2. f is contin- uous since it is a polynomial, and f (0) = 2, f (1) = 18, so f (0) < 5 < f (1). Multiply (2 x)(2 + x) (x 4)(2 + x) (2 + x) Therefore, lim x 4 the conjugate: (x 4)(2 + x) 4: d since x 3 , |x 3| = (x 3), so. |x 3| lim x 3 (x 1)(x 3) x2 4x + 3: c since the slope of the line is. , the vec- tor h3, 4iis parallel to the line, so the vector. 2h3, 4i = h 6, 8iis parallel to the line. 3: d since the numerator approaches 5 and the denominator approaches 0, we analyze the signs: as x 0 , x 5 < 0, x < 0, and x +