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10FallHwk03_Soln

# 10FallHwk03_Soln - Math 2413 Homework Set 3 Solutions z 6 5...

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Math 2413 Homework Set 3 – Solutions 10 Points 2. (2 pts) ( )( ) ( ) ( ) 2 6 6 10 10 5 4 3 4 2 5 4 6 5 4 lim lim lim lim 10 6 10 6 6 6 6 z z z z z z z z z z z z z z z z z z fi¥ fi¥ fi¥ fi¥ - - - + - - - - = = = = = -¥ + + + + 5. (2 pts) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 7 2 7 2 2 7 7 14 14 2 14 lim lim lim Assume 0 because so 6 6 7 6 14 14 14 lim lim 6 6 6 y y y y y y y y y y y y y y y y y y y y y y y y y fi¥ fi¥ fi¥ fi¥ fi¥ + + + = = > fi ¥ = + + + + + = = = + + For the second limit the work will be identical until we get rid of the absolute value. In this case, we can assume that 0 y < because y fi -¥ and so y y = - . So, picking up we get, ( ) 2 2 2 2 2 7 7 14 14 2 14 14 lim lim lim 6 6 6 6 7 y y y y y y y y y y y fi¥ fi¥ fi¥ + + + = = = - - + - + + 7. (2 pts) Not much to do here. The function is continuous because it is a difference/product of continuous functions. Now all we need to do is evaluate the function at the two points. ( ) ( ) 15 11.7923 7 4.0547 f f - = - = - So, we can see that ( ) ( ) 7 4.0547 0 15 11.7923 f f - = - < < - = and so, by the IVT there is a number c such that 15 7 c - < < - such that ( ) 0 f c = . Or in other words, c is a root of the function. For the sake of completeness we can use a computer to see that 11.48494037 x = - is a root of the function.

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