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

  1. Aug 24, 2010 #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. Aug 25, 2010 #2


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    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. Aug 25, 2010 #3
    I have no idea what you are talking about, I am sorry....
  5. Aug 25, 2010 #4
    What I have done:


    This is the inverse formula. What should I do now?
  6. Aug 25, 2010 #5


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    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. Aug 25, 2010 #6


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    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. Aug 25, 2010 #7


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    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. Aug 25, 2010 #8
    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. Aug 25, 2010 #9
    OK, I think I got it. Please check if I am right,

    Attached Files:

  11. Aug 25, 2010 #10


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

    so rearranging gives
    [tex] x(y) = \sqrt{\frac{1}{y}}[/tex]
  12. Aug 25, 2010 #11


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    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. May 17, 2011 #12
    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. May 17, 2011 #13


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    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. May 18, 2011 #14


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    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. May 18, 2011 #15


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    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. Dec 13, 2013 #16
    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}
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