MATH 151 Midterm: MATH 151 TAMU Y2013 2013a Exam 1a Solutions
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Exam i version a solutions: c let f (x) = 4x2 + 16x 2. f is contin- uous since it is a polynomial, and f ( 1) = 14, f (0) = 2, so f ( 1) < 5 < f (0). Therefore, by the intermediate value the- orem, there is a solution to f (x) = 5 on. [ 1, 0]: a from the graph of f shown below, we see that f is continuous, but not di eren- tiable at x = 3. 4: b the line tangent to the position graph at t = 2. 5 is positive, so the velocity (deriva- tive of position) is positive. v > 0, 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. Multiply (x 4)(2 + x) (2 x)(2 + x)
