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Andrew | Andrew | ||
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| + | '''Prof. Alekseenko:''' Let's see. The #8a says: Let <math>f(x)=\frac{1}{x^2}</math>, | ||
| + | <math>x\neq 0\,</math>. Determine the direct image <math>f(E)\,</math>, if | ||
| + | <math>E:=\{ x \in R:\, 1\le x \le 2 \} </math>. | ||
| + | |||
| + | The starting point is to understand what needs to be determined. In this problem, we need to find the "direct image". A reasonable starting point then is to recall the definition of the "direct image". | ||
| + | |||
| + | Andrew or somebody else: can you please find in the book and type here the definition | ||
| + | of the "direct image"? | ||
| + | |||
| + | Also, please see the solution below submitted by one of the students. Notice that this solution gives the correct answers. However, we may want to add some lines to explain what was the line of thought. (This solution will be just fine on the test, but for the discussion purposes, I think it will not hurt to add some detail.) | ||
| + | |||
| + | Believe me or not, in both parts (a) and (b) it would be good to start from recalling the corresponding definitions. | ||
---- | ---- | ||
8.(a) | 8.(a) | ||
| − | <math>f(1) = 1 / 1^2 = 1 / 1 = 1</math> | + | <math>f(1) = 1 / 1^2 = 1 / 1 = 1\,</math> |
| − | <math>f(2) = 1 / 2^2 = 1/4 </math> | + | <math>f(2) = 1 / 2^2 = 1/4\, </math> |
| − | <math>f(E) = | + | <math>f(E) = \{f(x)\, : \,1 \le x \le 2 \} </math> |
| − | <math> 1/4 \le f(E) \le 1</math> | + | <math> 1/4 \le f(E) \le 1 </math> |
| + | Then, <math> f(E) = \{ y : 1/4 \le y \le 1 \} </math> | ||
| − | + | <math> = [ 1/4, 1 ]\, </math> | |
| + | (b) | ||
| + | <math> f(1) = 1 / 1^{2} = 1 / 1 = 1 \, </math> | ||
| − | ( | + | <math> f(2) = 1 / 4^2 = 1/16 \, </math> |
| + | |||
| + | |||
| + | <math> f^{-1}(G) = \{ x \in R, 1 \le f(x) \le 4 \, : \, f(x) \in G \} </math> | ||
| + | |||
| + | <math>=\{ x : -1/2 \le x \le -1\ \mbox{or} \ 1/2 \le x \le 1 \} </math> | ||
| − | <math> | + | <math>= [-1, -1/2] \bigcup [1/2, 1] </math> |
| − | + | A Packers fan in the class | |
Latest revision as of 17:29, 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
Prof. Alekseenko: Let's see. The #8a says: Let $ f(x)=\frac{1}{x^2} $, $ x\neq 0\, $. Determine the direct image $ f(E)\, $, if $ E:=\{ x \in R:\, 1\le x \le 2 \} $.
The starting point is to understand what needs to be determined. In this problem, we need to find the "direct image". A reasonable starting point then is to recall the definition of the "direct image".
Andrew or somebody else: can you please find in the book and type here the definition of the "direct image"?
Also, please see the solution below submitted by one of the students. Notice that this solution gives the correct answers. However, we may want to add some lines to explain what was the line of thought. (This solution will be just fine on the test, but for the discussion purposes, I think it will not hurt to add some detail.)
Believe me or not, in both parts (a) and (b) it would be good to start from recalling the corresponding definitions.
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 \} $
$ = [ 1/4, 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\ \mbox{or} \ 1/2 \le x \le 1 \} $
$ = [-1, -1/2] \bigcup [1/2, 1] $
A Packers fan in the class
