MATH 151 Midterm: MATH 151 TAMU Y2014 2014c X2B Solutions

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31 Jan 2019
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Fri, 31/oct c(cid:13)2014 art belmonte f (x) f (a: (b) two limit de nitions of f (a) are lim x a f (a + h) f (a) and lim h 0. For f (x) = (2x + 1)3/2 and a = 1 , these are represented by choices (iii) and (ii), respectively. The other is rubbish. h x a: (a) for x 6= 0, f (x) = 5x4 + 1, (a) now v (t) = s (t) = (cid:0)t 2 + 1(cid:1) (1) t (2t) 25 ft/s: (d) the slope of the tangent is f (x) = 3x2 4x. So f (x) = 1 yields 0 = 3x2 4x + 1 = (3x 1) (x 1) whence x = 1. 3(cid:0)x2 1(cid:1) 2/3 (2x): (e) for x 6= 1, we have g (x) = 1 g (x) g (1) x 1. = shows that g (1) does not exist.

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