Homework 4-solutions

Homework 4-solutions - Version 028 – Homework 4 –...

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Unformatted text preview: Version 028 – Homework 4 – Helleloid – (58250) 1 This print-out should have 21 questions. Multiple-choice 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 , − 5) uniondisplay ( − 5 , 2) uniondisplay (2 , 10) 3. ( − 10 , 2) uniondisplay (2 , 10) 4. ( − 10 , − 5) uniondisplay ( − 5 , 10) 5. ( − 10 , − 5] uniondisplay [2 , 10) Explanation: Since f and g are piecewise linear, they are continuous individually on ( − 10 , 10) except at their ‘jumps’; i.e. , at x = 2 in the case of f and x = 2 , − 5 in the case of g . But the product of continuous functions is again continuous, so fg is certainly continuous on ( − 10 , − 5) uniondisplay ( − 5 , 2) uniondisplay (2 , 10) . The only question is what happens at x = 2 , − 5. 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 = 2, lim x → 2 − { f ( x ) g ( x ) } = − 10 = f (2) g (2) = lim x → 2+ { f ( x ) g ( x ) } , while at x = − 5, lim x →− 5 − { f ( x ) g ( x ) } = 0 = f ( − 5) g ( − 5) = lim x →− 5+ { f ( x ) g ( x ) } . Thus, fg is continuous at x = 2 and at x = − 5. Consequently, the product fg is continuous at all x in ( − 10 , 10) . 002 10.0 points Determine which of the following could be the graph of f near the origin when f ( x ) = x 2 − 7 x + 10 2 − x , x negationslash = 2 , 2 , x = 2 . 1. Version 028 – Homework 4 – Helleloid – (58250) 2 2. 3. 4. 5. 6. correct Explanation: Since x 2 − 7 x + 10 2 − x = ( x − 2)( x + 5) 2 − x = 5 − x, for x negationslash = 2, we see that f is linear on ( −∞ , 2) uniondisplay (2 , ∞ ) , 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 ) . Consequently, Version 028 – Homework 4 – Helleloid – (58250) 3 must be the graph of f near the origin. 003 10.0 points Find all values of x at which the function f defined by f ( x ) = x − 7 x 2 − 5 x + 6 is continuous, expressing your answer in in- terval notation. 1. ( −∞ , − 2) ∪ ( − 2 , 3) ∪ (3 , ∞ ) 2. ( −∞ , − 3) ∪ ( − 3 , − 2) ∪ ( − 2 , ∞ ) 3. ( −∞ , 3) ∪ (3 , ∞ ) 4. ( −∞ , 2) ∪ (2 , 3) ∪ (3 , ∞ ) correct 5. ( −∞ , 2) ∪ (2 , ∞ ) Explanation: After factorization the denominator be- comes x 2 − 5 x + 6 = ( x − 2)( x − 3) , so f can be written as f ( x ) = x − 7 ( x...
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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.

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Homework 4-solutions - Version 028 – Homework 4 –...

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