MATH 151 Midterm: MATH 151 TAMU Y2015 2015a X1B Solutions
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Fri, 20/feb c(cid:13)2015 art belmonte: (b) points a (1, 3) and b ( 5, 8) are on the line. A direction vector for the line is v = b a or [ 6, 11]. Thus [x(t), y(t)] = l (t) = a + tv is a vector equation of the line. Expansion gives x = 1 6t, y = 3 + 11t: (d) for f (x)= 4 (x 2) (x 1) (x 1) x. , lim x 1 f (x) = 4 and f (x) = . 1 x: (d) we have = bac = cos 1 ab ac. 4+1 9+25(cid:17) = cos 1(cid:16) 11 170(cid:17): (b) the tangent line to f (x) at x = 5 is y = 7. Hence its slope lim x 5 f (x) f (5) x 5 is m = 7. 2 : (a) now v = ab = b a = [ 2, 7] [1, 6] = [ 3, 1].
