Mac2311Q1B

Mac2311Q1B - you may try to ﬁnd this solution by using a...

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Solutions to MAC 2311 Quiz 1 B Jason Harrington January 28, 2006 1. (1 pt) Determine whether the statement is TRUE or FALSE: ”The range of f - 1 = domain of f This statement above is true. 2. (3 pts) Let y = sin( x ), write the equation of the graph that results from (a) shifting down 2 units y = sin( x ) + 2 (b) shifting 2 units to the right y = sin( x - 2) (c) reﬂecting about the y-axis f ( x + 2) = sin( - x ) 3. (2 pts) Find the inverse for the following function: f ( x ) = 5 1 - 3 x x = 5 1 - 3 y x 5 = 1 - 3 y x 5 - 1 = - 3 y 1 - x 5 = 3 y ln(1 - x 5 ) = ln(3 y ) ln(1 - x 5 ) = y ln(3) ln(1 - x 5 ) ln(3) = y Therefore: f - 1 ( x ) = ln(1 - x 5 ) ln(3) 4. (2 pts) Simplify the expression: sin(arctan x ) Let θ = arctan( x ). Now recall that - π 2 < θ < π 2 Now recall that: 1 + cot 2 ( θ ) = csc 2 ( θ ) so csc( θ ) = q 1 + cot 2 ( θ ) = p 1 + 1 /x 2 1

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= p 1 + 1 /x 2 * x 2 x 2 = x 2 + 1 x Hence, sin arctan( x ) = sin( θ ) = 1 csc( θ ) = 1 x 2 +1 x = x x 2 +1 Instead of using trignometric identities like I have done (ugh.
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Unformatted text preview: ..) you may try to ﬁnd this solution by using a diagram (sigh of relief). 5. (2 pts) Find the domain (in interval notation) for the following functions: (a) f ( t ) = √ 1-2 t Since we cannot take the square root of a negative number then: 1-2 t ≥ 0 so 1-2 t ≥ 1 ≥ 2 t ln(1) ≥ ln(2 t ) ≥ t ln(2) ≥ t Thus, the doman is (-∞ , 0] (b) g ( x ) = sin( x ) 2+cos( x ) We have no restrictions for the numerator so we just have to check the denominator. Is there are a value such that 2 + cos ( x ) = 0? Well. .. that would mean that cos( x ) =-2 and that never happens with real values. So the domain is the whole real line. Thus, the domain of g is (-∞ , ∞ ). 2...
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Mac2311Q1B - you may try to ﬁnd this solution by using a...

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