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Q1Solutions

# Q1Solutions - f The solutions are x = ± 1 To achieve x 2-1...

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MAC 2233 Quiz 1 Solutions 1. Find the equation of the line that passes through the point (2 , 4) and is perpendicular to the line 3 x + 4 y - 22 = 0. SOLUTION Solve the given equation for y to identify the slope of the given line. The slope of the given line is - 3 4 , so the slope of the line we want to ﬁnd is m = 4 3 . Now, we have both the slope and a point on the line whose equation we want to ﬁnd, so we should use the point-slope form of the equation of a line. The point-slope form of of a line is y - y 1 = m ( x - x 1 ) . (1) Using ( x 1 ,y 1 ) = (2 , 4) and m = 4 3 , equation (1) becomes y - 4 = 4 3 ( x - 2) . (2) Finally, simplify equation (2) to get y = 4 3 x + 4 3 . 2. Find the domain of the function f ( x ) = 1 x 2 - 1 . SOLUTION We need to consider two things here. First, we need to avoid divi- sion by zero (i.e. x 2 - 1 6 = 0) and second, we need to ensure that x 2 - 1 0 so that the quantity under the square root is not negative. To achieve x 2 - 1 6 = 0, simply solve the equation x 2 - 1 = 0 for x and exclude the solutions from the domain of

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Unformatted text preview: f . The solutions are x = ± 1. To achieve x 2-1 ≥ 0, ﬁrst observe that x 2-1 = 0 has solutions x = ± 1, and that we have already excluded these x ’s, so we need to ensure x 2-1 > 0. The easiest way to do this is to factor: x 2-1 = ( x + 1)( x-1) . From this factorization, we see that in order for x 2-1 > 0, we need x +1 and x-1 to have the same signs (i.e. either both should be negative or both should be positive). Both x + 1 and x-1 are negative when x <-1. Both x + 1 and x-1 are positive when x > 1. Therefore, the domain is (-∞ ,-1) ∪ (1 , ∞ ) . 3. Given the functions f ( x ) = 1 √ 2 x +1 and g ( x ) = √ x , ﬁnd ( f ◦ g )( x ). SOLUTION There is not much to do here. Simply rewrite f with x replaced by g ( x ) = √ x . The solution is ( f ◦ g )( x ) = f ( g ( x ) ) = 1 p 2 √ x + 1 ....
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Q1Solutions - f The solutions are x = ± 1 To achieve x 2-1...

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