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Mac2311Q1A

# Mac2311Q1A - θ = √ 1 x 2 Hence cos arctan x = cos θ = 1...

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

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Unformatted text preview: θ ) = √ 1 + x 2 . Hence, cos arctan( x ) = cos( θ ) = 1 sec( θ ) = 1 √ 1+ x 2 Instead of using trignometric identities like I have done (ugh. ..) you may try to ﬁnd this solution by using a diagram (sigh of relief). 1 5. (2 pts) Find the domain (in interval notation) for the following functions: (a) f ( x ) = e x 1+ e x Since 1 + e x 6 = 0 for all x , the domain is the whole real line. Or (-∞ , ∞ ) (b) g ( t ) = √ 1-3 t We cannot take the square root of a negative number i.e., 1-3 t ≥ so. .. 1-3 t ≥ 1 ≥ 3 t ln(1) ≥ ln(3 t ) ≥ t ln(3) ≥ t Thus, the doman is (-∞ , 0] 2...
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