1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse of a function

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex]
    f(x) = \sqrt{5x+2}
    [/tex]

    is one-to-one.

    2. Relevant equations

    If f' >0 or f' < 0 everywhere on f's domain, f is one-to-one.

    3. The attempt at a solution

    [tex]
    f^\prime = \frac{5}{2} \frac{1}{\sqrt{5x +2}} = \frac{5}{2f}
    [/tex]

    f' is positive on (-2/5, infinity) but is undefined at x = -2/5. I can therefore say that f is one-to-one on this interval, but what about the point x=-2/5? The derivative is undefined there (the left-hand limit does not exist). There might be a way to argue that f is increasing on [-2/5, infinity), but I don't know how to do so since I really don't know if f(-2/5) = 0 is the minimum value.
     
  2. jcsd
  3. Apr 7, 2015 #2

    joshmccraney

    User Avatar
    Gold Member

    You have shown that ##f## is monotonically increasing on ##(-2/5,\infty)##. This combined with the continuity of ##f## along ##x \in [0,\infty)## is sufficient to state that ##f## is one-to-one.

    Also, you can look at the definition of one-to-one when dealing with ##x=-2/5## (since you used a theorem for all other points of existence), and recall that one-to-one functions never map distinct elements of its domain to the same element of its range.
     
  4. Apr 7, 2015 #3
    Thanks! But without doing anything else, I'm not sure I can assume f(x) will never be zero except at x=-2/5.

    But after thinking a bit more, I was able to use the fact that f(a) = f(-2/5) is defined and the mean value theorem to write

    f(x) - f(a) = f'(c)(x - a) where c is in (a, x). Since f'(c) > 0 when x > a, the right-hand side will be positive and so f(x) > f(a) everywhere in [a, x) (with x > a).
     
  5. Apr 7, 2015 #4

    joshmccraney

    User Avatar
    Gold Member

    Very nice. You are in analysis, correct, hence the burden of proof?
     
  6. Apr 7, 2015 #5
    This was a problem in my calculus textbook. There's a theorem that says increasing/decreasing functions have inverses, but I wasn't sure how to show that the function satisfied the hypothesis.
     
  7. Apr 7, 2015 #6

    joshmccraney

    User Avatar
    Gold Member

    I see. I suppose another way to solve could have been to simply set ##f=0## and solve for ##x##. You'd then know what values of ##x## give the desired output. It seems then your theorem covers all other points.
     
  8. Apr 7, 2015 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Directly from the definition of "one-to-one": If f(x)= f(y) then [itex]\sqrt{5x+ 2}= \sqrt{5y+ 2}[/itex]. Show that x= y.
     
    Last edited by a moderator: Apr 7, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inverse of a function
  1. Inverse of a function (Replies: 0)

  2. Inverse of a function (Replies: 2)

  3. Inverse functions (Replies: 5)

  4. Inverse of a function (Replies: 2)

  5. Inverse Functions (Replies: 19)

Loading...