| Line 35: | Line 35: | ||
| − | <math>f^{-1}(G) = </math>{<math>x \in R, 1 | + | <math>f^{-1}(G) = </math>{<math>x \in R, 1 \le f(x) \le 4 : f(x) \in G</math>} |
= {<math>x : -1/2 \le x \le -1</math> or <math>1/2 \le x \le 1</math>} | = {<math>x : -1/2 \le x \le -1</math> or <math>1/2 \le x \le 1</math>} | ||
<math>= [-1, -1/2] \bigcup [1/2, 1] | <math>= [-1, -1/2] \bigcup [1/2, 1] | ||
Revision as of 13:03, 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 $
$ f^{-1}(G) = ${$ x \in R, 1 \le f(x) \le 4 : f(x) \in G $}
= {$ x : -1/2 \le x \le -1 $ or $ 1/2 \le x \le 1 $}
$ = [-1, -1/2] \bigcup [1/2, 1] $
