MATH 151 Midterm: MATH 151 TAMU Y2013 2013a Exam 3a Solutions
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Exam iii version a solutions: b switch x and y and solve for y: x = 2xy + 7x = 4y 1, 2xy 4y = 7x 1, y = 7x 1. 0, x = 1, and x = 2. Testing the signs on each subinterval, we nd that f > 0 on ( , 0) (0, 1) (2, ) and f < 0 on (1, 2). Therefore, f has a relative minimum at x = 2 only: e g (2) = Since g = f 1, if y = g(2), then f (y) = 2, which means y = 5. Using properties (e3 ln x)(ln(e2x)) (x3)(2x) = 2x4. of logarithms, eln(x3)(ln(e2x)): e we cannot simply cancel since. 2: c using properties of logarithms, f (x) = ln(x4)+ln(ex) = 4 ln x+x, so f (x) = 3 the range of f (x) = arcsin x. sin: d let y = arccos(cid:18) 1 triangle below, tan y = 15.
