M151 Practice Exam 1 soln

# M151 Practice Exam 1 soln - Math 151 PRACTICE EXAM 1...

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PRACTICE EXAM 1: Solutions Page 1 of 7 1. (15 total points) Solve each of the following equations. (a) [5 points] 3 x + 5 x + 2 = 2 Solution: x ( x + 2) · ± 3 x + 5 x + 2 ² = 2 · x ( x + 2) 3 x + 6 + 5 x = 2 x 2 + 4 x 0 = 2 x 2 - 4 x - 6 0 = x 2 - 2 x - 3 0 = ( x - 3)( x + 1) x = 3 or x = - 1 Both of these values are in the domain of the original equation. So the solutions are x = 3 and x = - 1 . (b) [5 points] e x 2 - 1 - 1 = 0 Solution: e x 2 - 1 = 1 x 2 - 1 = ln(1) x 2 - 1 = 0 x = ± 1 So x = ± 1. (c) [5 points] log(2 - x ) = log( x - 2) + log(2 x + 1) Solution: log(2 - x ) = log(( x - 2)(2 x + 1)) 2 - x = ( x - 2)(2 x + 1) 2 - x = 2 x 2 - 3 x - 2 0 = 2 x 2 - 2 x - 4 0 = x 2 - x - 2 0 = ( x - 2)( x + 1) x = 2 or x = - 1 Neither of these values is in the domain of the original equation. So there are no solutions.

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PRACTICE EXAM 1: Solutions Page 2 of 7 2. (10 total points) Sketch the graph of each of the following functions. (a) [5 points] f ( x ) = - x 2 - 4 x - 5 Solution: f ( x ) = - ( x 2 + 4 x ) - 5 = - ( x 2 + 4 x + 4) - 5 + 4 = - ( x + 2) 2 - 1 So the vertex is at ( - 2 , - 1). Since the parabola opens downward, there are no x - intercepts. The y -intercept is at (0 , - 5). x y 5 - 5 (b) [5 points] g ( x ) = 2 sin ( πx + π ) - 3 Solution: g ( x ) = 2 sin( π ( x + 1)) - 3 The amplitude is 2. The period is 2. The phase shift is - 1. The vertical shift is - 3. x
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M151 Practice Exam 1 soln - Math 151 PRACTICE EXAM 1...

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