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Determine Direct and Inverse Image f(E) and f^1 (G)... 
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#1
Aug2410, 11:21 PM

P: 157

1. The problem statement, all variables and given/known data
Let f(x):=1/x^2, x not equal 0, x belongs R a) Determine the direct image f(E) where E:= (x belongs R : 1<=x<=2) b) Determine the inverse image f^(1)(G) where G:= (x belongs R : 1<=x<=4) 2. Relevant equations 3. The attempt at a solution A) Let f: R > R be defined by f(x):=1/x^2. Then, the direct image of the set E:=(x:1<=x<=2) is the set f(E)=(y:1<=x<=1/4). If G:= (y : 1<=x<=4), then the inverse image of G is the set f^1 (G)=(x: And here I don't quite understand how to find an inverse image. Please help. 


#2
Aug2510, 12:18 AM

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P: 3,307

how about starting by finding the inverse map...
if you choose a reasonable domain f will be 1:1, so its inverse will exist such that f^(1)(f(x)) = x 


#3
Aug2510, 12:36 AM

P: 157

I have no idea what you are talking about, I am sorry....



#4
Aug2510, 01:07 AM

P: 157

Determine Direct and Inverse Image f(E) and f^1 (G)...
What I have done:
f^(1)(x)=x^2 This is the inverse formula. What should I do now? 


#5
Aug2510, 01:56 AM

HW Helper
P: 3,307

I prefer to keep the variables differnt for the function & its inverse as below, i find it makes thing easier, so let:
[itex] y = f(x) [/itex] is a function that maps from x to y for question a), if the domain of x is [1,2], then the y is in the range [1/4,1] as you've found now for question b) assuming f is 1:1 we can find its inverse function, lets call it g [itex] x = f^{1}(y) = g(y) [/itex] which maps y back to x 


#6
Aug2510, 02:01 AM

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P: 3,307

now with all that in mind, starting with the function
[tex] y = f(x) = \frac{1}{x^2}[/tex] you've said the inverse funtion is [tex] x = g(y) = f^{1}(y) = y^2[/tex] lets check if it actually satisfies the inverse property [tex] f^{1}(f(x)) = g(f(x)) = (f(x))^2 = (\frac{1}{x^2})^2 = \frac{1}{x^4} \neq x [/tex] so this is not infact the correct inverse function 


#7
Aug2510, 02:02 AM

HW Helper
P: 3,307

to find the correct inverse, start with
[tex] y = \frac{1}{x^2}[/tex] now solve for x in terms of y, that gives you g 


#8
Aug2510, 09:56 AM

P: 157

I am so lost with your last comment. I can only see that this formula is the same as in your previous post.
Please any clues? 


#9
Aug2510, 10:13 AM

P: 157

OK, I think I got it. Please check if I am right,



#10
Aug2510, 10:47 AM

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P: 3,307

yep that looks good
[tex] y(x) = \frac{1}{x^2}[/tex] so rearranging gives [tex] x(y) = \sqrt{\frac{1}{y}}[/tex] 


#11
Aug2510, 01:04 PM

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PF Gold
P: 39,345




#12
May1711, 09:55 AM

P: 157

Hi! This is an old posting of mine. I am going back to this discussion. Is it right that for question b the answer is:
f^(1)(G)={1/2<=x<=1} ? Thanks! 


#13
May1711, 11:59 AM

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P: 39,345

No, it's not. Notice that if x= 1/2, y= 1/(1/2)^= 4. f is an even function so its graph is symmetric about the yaxis. lanedance made an error when he said the inverse function is given by [itex]x= 1/\sqrt{y}[/itex]. In fact, this function does not have an inverse!
In my first year in grad school, I had to present, before the class, a proof involving [itex]f^{1}(A)[/itex] where A is a set. I did the whole problem assuming that f had an inverse very embarassing! The way you should have approached the problem is to look for x such that f(x)= 1/x^2= 1 and f(x)= 1/x^2= 4. The first equation has solutions 1 and 1, the second has solutions 1/2 and 1/2. Further, the derivative of f is 1/x^3 which is always negative between 1 and 2 and positive between 2 and 1. The fact that the function is decreasing on one interval and negative on the other tells you the domain is just the two intervals, [itex]1< x\le 2[/itex] and [itex]2\le x< 1[/itex]. If, for some x in either of those intervals, the derivative were 0, you would have to consider the possibility that there is a turning point in the interval. 


#14
May1811, 01:27 PM

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P: 3,307

this isn't very fresh in my head, but I'm not sure I understand
Unless I'm missing something f(x):=1/x^2 on 1<=x<=4 is monotone decreasing with f'(x) not zero anywhere so the inverse exist 


#15
May1811, 03:38 PM

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P: 39,345

But it is not x, in the original function, that is between 1 and 4. The problem was to find [itex]f^{1}(G)[/itex] where G is [1, 4]. If we write the original function [itex]y= 1/x^2[/itex]. Then we are looking for x such that [itex]1\le y\le x[/itex].
Yes, the problem was written in terms of [itex]f^{1}(x)[/itex] with [itex]1\le x\le 4[/itex] but in terms of the original function, y= f(x), it is y that is between 1 and 4. Is there a reason why LaTeX is not working? 


#16
Dec1313, 01:26 AM

P: 1

The solution of f(E) is not right because 1≤ y≤ 1/4 would imply the set (∞,1/4] U [1,∞). But, f(1.5) = 0.444 ∉ (∞,1/4] U [1,∞) whereas 1.5 satisfies x ∈ [1,2]. Therefore f(E) is derived as follows:
f(E)= f{x∈ ℝ : 1≤ x ≤ 2} Now, 1≤ x ≤ 2 ⇔ 1≥ 1/x ≥ 1/2 ⇔ 1≥ 1/x^2 ≥ 1/4 ⇔ 1≥ y ≥ 1/4 ∴ f(E) = {y∈ ℝ : 1/4 ≤ y ≤ 1} 


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