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Then, <math>f(E) = </math>{<math>y : 1/4 \le y \le 1</math>}
 
Then, <math>f(E) = </math>{<math>y : 1/4 \le y \le 1</math>}
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(b)
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<math>f(1) = 1 / 1^2 = 1 / 1 = 1</math>
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<math>f(2) = 1 / 4^2 = 1/16 </math>

Revision as of 12:48, 7 February 2010

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I am having trouble with question 8 from our first HW. I am going back and trying to fix my old HW. I got back my HW and got a 0 out of 25. Can any one help get me started?

Thanks,

Andrew


8.(a)

$ f(1) = 1 / 1^2 = 1 / 1 = 1 $

$ f(2) = 1 / 2^2 = 1/4 $


$ f(E) = ${$ f(x) : 1 \le x \le 2 $}

$ 1/4 \le f(E) \le 1 $


Then, $ f(E) = ${$ y : 1/4 \le y \le 1 $}


(b)

$ f(1) = 1 / 1^2 = 1 / 1 = 1 $

$ f(2) = 1 / 4^2 = 1/16 $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett