This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ) ....
View
Full
Document
This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
 schultz
 Differential Calculus

Click to edit the document details