Exam1.solutions

Exam1.solutions - Exam 1 Solutions, Summer '08, 1. What is...

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Exam 1 Solutions, Summer ’08,  1 . What is the domain of 1 ) ln( 4 ) ( - + - = x x x x f ?  Express your answer using interval   notation.           Numerator:  domain of  ln (x) is  ) , 0 ( (the log function by definition can only be applied  to positive numbers); to make sure we’re not taking the square root of a negative  number, we require:  ] 4 , ( 4 0 4 -∞ - x x Denominator:   ) , 1 ( ) 1 , ( 1 -∞ x Combining these implies the domain of f(x) is  ] 4 , 1 ( ) 1 , 0 (      2.  Suppose  f(x) =  x 3  + 2 + 5sin(x 2  + 4x) a . What is the sign of  f(2)?  What is the sign of  f(- 2)? f (2) = 10 + 5sin(12).   Since  1 ) sin( 1 - x for all x, we have that,  5 ) 12 sin( 5 5 - So, it must be that:     5 10 ) 2 ( 5 10 + - f  , i.e. f(2) must be between 5 and 15, which  implies f(2) is positive. f(-2) = -6 + 5sin(-4) and by the same reasoning as above, we have that:   5 6 ) 2 ( 5 6 + - - - - f  which implies that f(-2) is negative. b . Explain why the equation  x 3  + 2 + 5sin(x 2  + 4x )=  0  must have at least   one solution.  Give an interval in which the solution must be found. You must quote a specific result  from the course and explain its relevance (your answers to part  a.  may be useful). f(x) is the sum of continuous functions so it’s continuous for all x, and in  particular, it’s continuous on the interval [-2, 2 ]. Therefore, according to the  Intermediate Value Theorem, the equation f(x) = 0 must have at least one solution in the  interval (-2, 2 ) since f(-2) and f(2) are of opposite signs.
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  3 .  A portion of the graph of a function,  f(x)  is shown below. Use the graph to answer  the questions that follow.                                                    x y   a . For which values of  x,  if any in the interval   [-6,6],   is  f(x)  not continuous? Explain your  answer x = 5 since there is clearly a break in the graph at x = 5, (the left hand and right limits of 
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This note was uploaded on 08/10/2008 for the course MATH 135 taught by Professor Noone during the Spring '08 term at Rutgers.

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Exam1.solutions - Exam 1 Solutions, Summer '08, 1. What is...

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