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Unformatted text preview: Version 028 – EXAM 1 – Helleloid – (58250) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table t (min) 5 10 15 20 25 30 V (gal) 689 495 211 177 16 show the volume, V ( t ), of water remaining in the tank (in gallons) after t minutes. If P is the point (15 , V (15)) on the graph of V as a function of time t , find the slope of the secant line PQ when Q = (25 , V (25)). 1. slope = − 39 2. slope = − 28 . 4 3. slope = − 3 . 4 4. slope = − 47 . 8 5. slope = − 19 . 5 correct Explanation: When P = (15 , V (15)) , Q = (25 , V (25)) the slope of the secant line PQ is given by rise run = V (25) − V (15) 25 − 15 . From the table of values, therefore, we see that slope = 16 − 211 25 − 15 = − 19 . 5 . 002 10.0 points Below is the graph of a function f . 2 4 6 − 2 − 4 − 6 2 4 6 8 − 2 − 4 Use the graph to determine lim x → 4 f ( x ) . 1. limit = 8 2. limit = 6 3. limit = 7 4. limit does not exist correct 5. limit = 2 Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 8; and a right hand limit which is equal to 3. Since the two numbers do not coincide, the limit does not exist . 003 10.0 points Determine lim x → x − 1 x 2 ( x + 2) . 1. limit = 0 2. limit = ∞ 3. limit = − 1 2 4. none of the other answers 5. limit = −∞ correct Version 028 – EXAM 1 – Helleloid – (58250) 2 6. limit = 1 Explanation: Now lim x → x − 1 = − 1 . On the other hand, x 2 ( x + 2) > 0 for all small x , both positive and negative, while lim x → x 2 ( x + 2) = 0 . Consequently, limit = −∞ . keywords: evaluate limit, rational function 004 10.0 points Let F be the function defined by F ( x ) = x 2 − 64  x − 8  . Determine if lim x → 8 F ( x ) exists, and if it does, find its value. 1. limit = 16 2. limit = 8 3. limit = − 8 4. limit = − 16 correct 5. limit does not exist Explanation: After factorization, x 2 − 64  x − 8  = ( x + 8)( x − 8)  x − 8  . But, for x < 8,  x − 8  = − ( x − 8) . Thus F ( x ) = − ( x + 8) , x < 8 , By properties of limits, therefore, the limit exists and lim x → 8 F ( x ) = − 16 . 005 10.0 points Determine lim x → 6 x − 6 √ x + 3 − 3 . 1. limit = 3 2. limit = 1 3 3. limit doesn’t exist 4. limit = 1 6 5. limit = 6 correct Explanation: After rationalizing the denominator we see that 1 √ x + 3 − 3 = √ x + 3 + 3 ( x + 3) − 9 = √ x + 3 + 3 x − 6 ....
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This note was uploaded on 12/04/2009 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Calculus

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