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

  1. 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. jcsd
  3. lanedance

    lanedance 3,307
    Homework Helper

    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
  4. I have no idea what you are talking about, I am sorry....
  5. What I have done:


    This is the inverse formula. What should I do now?
  6. lanedance

    lanedance 3,307
    Homework Helper

    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
    Last edited: Aug 25, 2010
  7. lanedance

    lanedance 3,307
    Homework Helper

    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
  8. lanedance

    lanedance 3,307
    Homework Helper

    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
  9. 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?
  10. OK, I think I got it. Please check if I am right,

    Attached Files:

  11. lanedance

    lanedance 3,307
    Homework Helper

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

    so rearranging gives
    [tex] x(y) = \sqrt{\frac{1}{y}}[/tex]
  12. HallsofIvy

    HallsofIvy 41,051
    Staff Emeritus
    Science Advisor

    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}.

  13. 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:


    ? Thanks!
  14. HallsofIvy

    HallsofIvy 41,051
    Staff Emeritus
    Science Advisor

    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 [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.
  15. lanedance

    lanedance 3,307
    Homework Helper

    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
  16. HallsofIvy

    HallsofIvy 41,051
    Staff Emeritus
    Science Advisor

    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?
    Last edited by a moderator: Jun 25, 2011
  17. 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∈ : 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∈ : 1/4 ≤ y ≤ 1}
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted