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Homework Help: Analysis - Inverse Functions

  1. Sep 6, 2012 #1
    The problem statement, all variables and given/known data:
    Define f: ℝ→ℝ by f(x)=x^2. Find f^-1(T) for each of the following:

    (a) T = {9}
    (b) T = [4,9)
    (c) T = [-4,9]

    The attempt at a solution:
    So, the inverse of f should be f^-1(T)=+/-√(x). Therefor:

    (a) f^-1(9)= +/- 3
    (b) f^-1(4)= +/- 2, f^-1(5)= +/- √(5), f^-1(6)= +/- √(6), f^-1(7)= +/- √(7), f^-1(8)= +/- 2√(2)
    (c) Assuming I did the above correct, I have no idea how to do this part because clearly √(x) is not going to have a real solution from [-4,-1]

    Any help would be awesome! Thanks!
  2. jcsd
  3. Sep 6, 2012 #2


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    The function f(x) = x2, having the domain, , does not have an inverse function.

    What f -1(T) refers to is called the "inverse image" of set T for the function f.

    Does your textbook have a definition for the inverse image, f -1(T), where T is a set?
  4. Sep 6, 2012 #3
    They have one, but it's about 2 sentences long. If I'm understanding correctly:

    C [itex]\subseteq[/itex] A


    f: C → f(C) and, therefor

    f^-1: f(C) → f^-1(C)

    I might be abusing notation a bit on this, so please correct me.

    What I don't understand is how exactly do I find f^-1 if it is not the inverse function?
  5. Sep 7, 2012 #4


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    The very, very first proof I had to present before a class in graduate school had to do with "[itex]f^{-1}(X)[/itex]" for X a set. I did the whole proof assuming that f was invertible! Very embarassing!

    hamonjj, these are sets- otherwise "[itex]f^{-1}[/itex]" for [itex]f(x)= x^2[/itex] wouldn't make sense.

    The definition of [itex]f^{-1}(A)[/itex] for A a set is:
    [itex]f^{1}(A)= \{ x| f(x)\in A\}[/itex]. In particular, [itex]f^{-1}[/itex] of a set is a set. Yes, f(3)= 9 and f(-3)= 9 so that [itex]f^{-1}(9)= \{-3, 3\}[/itex]- be sure to write the answer as a set.

    Why are you looking at integers only? We are talking about a function from R to R, not integers. f(2)= 4 and f(3)= 9. And if 2< x< 3 then 4< x^2< 9. f of any number between 2 and 3 is in this set- the interval [2, 3) is in this set (do you see why 3 is NOT in the set?). But it is also true that f(-2)= 4 and f(-3)= 9 so the interval (-3, 2] is in the set. [itex]f^{-1}[4, 9)= [2, 3)\cup (-3, 2][/itex].

    For (c), there are NO (real) x such that f(x)< 0 so we can ignore the "-4" part. But any number from -3 to 3 will have square between 0 and 9 and so between -4 and 9.
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