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Inverse of a function

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

    Show that

    f(x) = \sqrt{5x+2}

    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

    f^\prime = \frac{5}{2} \frac{1}{\sqrt{5x +2}} = \frac{5}{2f}

    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


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    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


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    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


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    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


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    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
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