calculus test - exams Calculus PracticeTest Solutions Name:...

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Unformatted text preview: exams Calculus PracticeTest Solutions Name: 1. (3 points) Compute lim t 2 t 2- 4 t 2- 2 t + 1 . Solution Since the limit of the numerator is 0 and the limit of the denominator is not zero, we actually can use the plug-in theorem (aka substitution theorem). Thus we have lim t t 2- 4 t 2- 2 t + 1 = 1 = 0 . 2. (6 points) Let f ( x ) = 3 2 x- 1. Find the linearization L ( x ) to f ( x ) at x = 14. Then use this to approximate f (14 . 01). Solution The linearization is given by L ( x ) = f (14) + f (14)( x- 14), and since f (14) = 3, we have L ( x ) = 3 + f (14)( x- 14). It remains to find f (14). We have that f ( x ) = 1 3 (2 x- 1)- 2 3 2, so f (14) = 2 3 (27 2 3 ) = 2 27 . Thus L ( x ) = 3 + 2 27 ( x- 14). If we use L to approximate f when x = 14 . 01, we obtain f (14 . 01) L (14 . 01) = 3 + 2 27 1 100 = 3 + 1 1350 . 3. (5 points) Ship A leaves Bridgetown harbor and sails due west at 10 miles per hour. Ship B leaves from the same harbor one hour later, and sails due south at 12 miles per hour. If t represents elasped time since ship B left the harbor (measured in hours), find an expression for...
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calculus test - exams Calculus PracticeTest Solutions Name:...

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