# Exam3 9 - Find the value of f (ln 2) when f ( x ) = 2 e 2 x...

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Betancourt, Daniel – Exam 3 – Due: May 1 2007, 11:00 pm – Inst: Diane Radin 9 keywords: 015 (part 1 of 1) 10 points Determine f ( t ) when f 00 ( t ) = 2(6 t + 1) and f 0 (1) = 5 , f (1) = 6 . 1. f ( t ) = 6 t 3 + t 2 - 3 t + 2 2. f ( t ) = 2 t 3 - t 2 + 3 t + 2 3. f ( t ) = 6 t 3 + 2 t 2 - 3 t + 1 4. f ( t ) = 2 t 3 - 2 t 2 + 3 t + 3 5. f ( t ) = 2 t 3 + t 2 - 3 t + 6 correct 6. f ( t ) = 6 t 3 - 2 t 2 + 3 t - 1 Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 6 t 2 + 2 t + C where C is an arbitrary constant. But if f 0 (1) = 5, then f 0 (1) = 6 + 2 + C = 5 , i.e., C = - 3 . From this it follows that f 0 ( t ) = 6 t 2 + 2 t - 3 . The most general anti-derivative of f is thus f ( t ) = 2 t 3 + t 2 - 3 t + D , where D is an arbitrary constant. But if f (1) = 6, then f (1) = 2 + 1 - 3 + D = 6 , i.e., D = 6 . Consequently, f ( t ) = 2 t 3 + t 2 - 3 t + 6 . keywords: antiderivatives, second order derivatives, function value 016 (part 1 of 1) 10 points
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Unformatted text preview: Find the value of f (ln 2) when f ( x ) = 2 e 2 x , f (0) = 5 . 1. f (ln 2) = 9 2. f (ln 2) = 8 correct 3. f (ln 2) = 11 4. f (ln 2) = 12 5. f (ln 2) = 10 Explanation: Since d dx e αx = αe αx , we see that f ( x ) = e 2 x + C where the arbitrary constant C is speci±ed by the condition f (0) = 5. For then f (0) = 5 = ⇒ 1 + C = 5 , in which case f ( x ) = e 2 x + 4 . On the other hand, e 2 x f f f x =ln 2 = e 2 ln 2 = e ln 4 = 4 . Consequently, f (ln 2) = 4 + 4 = 8 . keywords: antiderivative, exponential func-tion, function value,...
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## This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas.

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