MATH 151 Midterm: MATH 151 TAMU Y2015 2015c X3Combo
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Sat, 05/dec c(cid:13)2015 art belmonte: (a) now f (x) = 2x 2x 2 = 0 gives x3 = 1 or. 1500 ln 3: (d) assuming exponential decay, the mass y at time t is y = y0ekt = 100ekt . When t = 350 years, y = 100(cid:0) 3: (a) now f (x) = 4 sin x 2 cos x ex +c. So c = 8 and therefore f ( ) = 0 + 2 e + 8 = 10 e . 10(cid:1)7/30 mg: (e) function f is concave up where f (x) > 0; i. e. , ( 6, 2) (8, ), (b) note 23x = 8x. 8x ln 8 5x ln 5: (c) now f (x) = 1 indeterminate quotient: lim x 0 ln(cid:0) 8. Hence the critical numbers of f are x = 1, 1: (e) with f (x) =(cid:0)x3 + 3x + 2(cid:1)1/2.
