# Determine Direct and Inverse Image f(E) and f^-1 (G)...

by phillyolly
Tags: determine, image, inverse
 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.
 HW Helper 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
 P: 157 I have no idea what you are talking about, I am sorry....
 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?
 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: $y = f(x)$ 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 $x = f^{-1}(y) = g(y)$ which maps y back to x
 HW Helper P: 3,307 now with all that in mind, starting with the function $$y = f(x) = \frac{1}{x^2}$$ you've said the inverse funtion is $$x = g(y) = f^{-1}(y) = y^2$$ lets check if it actually satisfies the inverse property $$f^{-1}(f(x)) = g(f(x)) = (f(x))^2 = (\frac{1}{x^2})^2 = \frac{1}{x^4} \neq x$$ so this is not infact the correct inverse function
 HW Helper P: 3,307 to find the correct inverse, start with $$y = \frac{1}{x^2}$$ now solve for x in terms of y, that gives you g
 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?
 P: 157 OK, I think I got it. Please check if I am right, Attached Thumbnails
 HW Helper P: 3,307 yep that looks good $$y(x) = \frac{1}{x^2}$$ so rearranging gives $$x(y) = \sqrt{\frac{1}{y}}$$
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PF Gold
P: 39,565
 Quote by phillyolly 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).
Everyone has been focusing on (B) but let me make an obvious point 1 is not "less than or equal to" 1/4! Also, it make no sense to say "y such that x has some property". What you should have is f(E)= {y: 1/4<= y<= 1}.

 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.
 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!
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 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 y-axis. lanedance made an error when he said the inverse function is given by $x= 1/\sqrt{y}$. 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 $f^{-1}(A)$ 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, $1< x\le 2$ and $-2\le x< -1$. 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.
 HW Helper 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
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 But it is not x, in the original function, that is between 1 and 4. The problem was to find $f^{-1}(G)$ where G is [1, 4]. If we write the original function $y= 1/x^2$. Then we are looking for x such that $1\le y\le x$. Yes, the problem was written in terms of $f^{-1}(x)$ with $1\le x\le 4$ 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?
 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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