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Unformatted text preview: white (taw933) – HW04 – benzvi – (55600) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Functions f and g are defined on ( − 10 , 10) by their respective graphs in 2 4 6 8 − 2 − 4 − 6 − 8 4 8 − 4 − 8 f g Find all values of x where the product, fg , of f and g is continuous, expressing your answer in interval notation. 1. ( − 10 , 10) correct 2. ( − 10 , − 4) uniondisplay ( − 4 , 10) 3. ( − 10 , 5) uniondisplay (5 , 10) 4. ( − 10 , − 4] uniondisplay [5 , 10) 5. ( − 10 , − 4) uniondisplay ( − 4 , 5) uniondisplay (5 , 10) Explanation: Since f and g are piecewise linear, they are continuous individually on ( − 10 , 10) except at their ‘jumps’; i.e. , at x = 5 in the case of f and x = 5 , − 4 in the case of g . But the product of continuous functions is again continuous, so fg is certainly continuous on ( − 10 , − 4) uniondisplay ( − 4 , 5) uniondisplay (5 , 10) . The only question is what happens at x = 5 , − 4. To do that we have to check if lim x → x − { f ( x ) g ( x ) } = f ( x ) g ( x ) = lim x → x + { f ( x ) g ( x ) } . Now at x = 5, lim x → 5 − { f ( x ) g ( x ) } = − 15 = f (5) g (5) = lim x → 5+ { f ( x ) g ( x ) } , while at x = − 4, lim x →− 4 − { f ( x ) g ( x ) } = 0 = f ( − 4) g ( − 4) = lim x →− 4+ { f ( x ) g ( x ) } . Thus, fg is continuous at x = 5 and at x = − 4. Consequently, the product fg is continuous at all x in ( − 10 , 10) . 002 10.0 points If the function f is continuous everywhere and f ( x ) = x 2 − 16 x − 4 when x negationslash = 4, find the value of f (4). 1. f (4) = 16 2. f (4) = − 4 3. f (4) = − 16 4. f (4) = − 8 5. f (4) = 8 correct 6. f (4) = 4 Explanation: white (taw933) – HW04 – benzvi – (55600) 2 Since f is continuous at x = 4, f (4) = lim x → 4 f ( x ) . But, after factorization, x 2 − 16 x − 4 = ( x − 4)( x + 4) x − 4 = x + 4 , whenever x negationslash = 4. Thus f ( x ) = x + 4 for all x negationslash = 4. Consequently, f (4) = lim x → 4 ( x + 4) = 8 . 003 10.0 points Determine which of the following could be the graph of f near the origin when f ( x ) = x 2 − x − 2 x − 2 , x negationslash = 2 , 2 , x = 2 . 1. correct 2. 3. 4. 5. 6. Explanation: Since x 2 − x − 2 x − 2 = ( x − 2)( x + 1) x − 2 = x + 1 , for x negationslash = 2, we see that f is linear on ( −∞ , 2) uniondisplay (2 , ∞ ) , white (taw933) – HW04 – benzvi – (55600) 3 while lim x → 2 f ( x ) = 3 negationslash = f (2) . Thus the graph of f will be a straight line of slope 1, having a hole at x = 2. This eliminates four of the possible graphs. But the two remaining graphs are the same except that in one f (2) > lim x → 2 f ( x ) , while in the other f (2) < lim x → 2 f ( x ) ....
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 Spring '08
 schultz
 Differential Calculus, lim, Continuous function, RLM

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