Q6Solutions - A P = ± 1 + r m ² mt ln ± A P ² = mt ln...

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MAC 2233 Quiz 6 Solutions July 24, 2009 1. Let f ( x ) = ( x - 1) e 3 x +2 . Determine the intervals where f is increasing. Solution: We want to determine where f 0 > 0. First, find f 0 : f 0 ( x ) = 1 · e 3 x +2 + ( x - 1) · 3 · e 3 x +2 = e 3 x +2 (1 + 3( x - 1)) = e 3 x +2 (3 x - 2) . Since e 3 x +2 > 0 for all x , and 3 x - 2 > 0 if and only if x > 2 3 , f is increasing on ( 2 3 , ). 2. Let g ( t ) = ln(ln( t 2 )). Find g 0 ( t ). Solution: This is an exercise in using the chain rule multiple times. g 0 ( t ) = 1 ln( t 2 ) · d dt ± ln( t 2 ) ² = 1 ln( t 2 ) · 1 t 2 · d dt ( t 2 ) = 1 ln( t 2 ) · 1 t 2 · 2 t = 2 t ln( t 2 ) = 2 2 t ln( t ) = 1 t ln( t ) . 3. Solve for t . A = P ± 1 + r m ² mt Solution: First, divide both sides by P , then take the natural logarithm of both sides.
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Unformatted text preview: A P = ± 1 + r m ² mt ln ± A P ² = mt ln ± 1 + r m ² . Then divide both sides by m ln ± 1 + r m ² to get ln( A P ) m ln(1 + r m ) = t. 4. Solve for r : A = Pe rt Solution: Try these steps on your own. First, divide both sides by P . Second, take the natural logarithm of both sides. Last, divide both sides by t . Your final solution should be 1 t ln ± A P ² = r....
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This note was uploaded on 12/27/2011 for the course MAC 2233 taught by Professor Smith during the Spring '08 term at University of Florida.

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Q6Solutions - A P = ± 1 + r m ² mt ln ± A P ² = mt ln...

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