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Final - Version 029 FINAL Helleloid(58250 This print-out...

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Version 029 – FINAL – Helleloid – (58250) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Below is the graph of a function f . -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 6 2 4 6 2 4 6 8 2 4 Use the graph to determine lim x → − 4 f ( x ). 1. lim x → − 4 f ( x ) = 6 correct 2. lim x → − 4 f ( x ) = 2 3. lim x → − 4 f ( x ) does not exist 4. lim x → − 4 f ( x ) = 10 5. lim x → − 4 f ( x ) = 8 Explanation: From the graph it is clear the f has both a left hand limit and a right hand limit at x = 4; in addition, these limits coincide. Thus lim x → − 4 f ( x ) = 6. 002 10.0 points Determine lim x 2 braceleftBig 1 x 2 2 x 2 2 x bracerightBig . 1. limit = 1 2 2. limit = 2 3. limit = 1 2 correct 4. limit does not exist 5. limit = 1 3 6. limit = 2 7. limit = 1 3 Explanation: After simplification we see that 1 x 2 2 x 2 2 x = x 2 x ( x 2) = 1 x , for all x negationslash = 2. Thus limit = lim x 2 1 x = 1 2 . 003 10.0 points Find all values of x at which the function f defined by f ( x ) = x 5 x 2 4 x + 3 is continuous, expressing your answer in in- terval notation. 1. ( −∞ , 3) ( 3 , 1) ( 1 , ) 2. ( −∞ , 1) (1 , 3) (3 , ) correct 3. ( −∞ , 1) ( 1 , 3) (3 , ) 4. ( −∞ , 1) (1 , ) 5. ( −∞ , 3) (3 , ) Explanation:
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Version 029 – FINAL – Helleloid – (58250) 2 After factorization the denominator be- comes x 2 4 x + 3 = ( x 1)( x 3) , so f can be written as f ( x ) = x 5 ( x 1)( x 3) . Being a rational function, it will be contin- uous everywhere except at the zeros of the denominator since it will not be defined at such points. Thus f is continuous everywhere except at x = 1 and x = 3. Hence it will continuous on ( −∞ , 1) (1 , 3) (3 , ) . 004 10.0 points If f is a function having -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 as its graph, which of the following is the graph of the derivative f of f ? 1. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 2. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 3. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 correct 4. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 5. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 Explanation: The tangent line to the graph of f is hor- izontal at ( 2 , f ( 2)) and (1 , f (1)), so the graph of f must have x -intercepts at x = 2 and x = 1; in particular, the graph of f can- not be a straight line. This eliminates three of the possible graphs. By observing the slope of the graph of f near x = 2 and x = 1 we see that the graph of f is
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Version 029 – FINAL – Helleloid – (58250) 3 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 2 4 2 4 2 4 2 4 005 10.0 points After t seconds the displacement, s ( t ), of a particle moving rightwards along the x -axis is given (in feet) by s ( t ) = 4 t 2 3 t + 6 .
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