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Functions 3

  1. Mar 1, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that the function [tex]f(x)=\sqrt(1-\sqrt(1-x^2))[/tex] has a finite one sided derivative at the point x=0.

    2. Relevant equations

    3. The attempt at a solution

    What the heck is a one sided derivative? If I differentiate it, and put limit x=0, I get infinity, which is definately NOT finite!

    Do I use the fundamental definition of a derivative here and put lim x=0- or something?
  2. jcsd
  3. Mar 1, 2007 #2


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    Yeah, prove that one of the following (or both of the following) is/are finite:

    [tex]\lim _{h \to 0^+} \frac{f(h) - f(0)}{h}[/tex]

    [tex]\lim _{h \to 0^-} \frac{f(h) - f(0)}{h}[/tex]

    Technically, the questions asks to show that f has a (as in, at least one) finite one-sided derivative, so you would only have to show that one of the above is finite. To be safe, however, you might want to show it for both.
  4. Mar 2, 2007 #3


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    As for the limit from the negative side, you wouldn't have any real number solutions.

    Knowing that f(0)=0, it is advantageous to study the squared fraction [tex](\frac{f(h)}{h})^{2}[/tex] in some detail..
  5. Mar 4, 2007 #4
    Thanks all, I solved it. Only one of the derivatives exists.
  6. Mar 5, 2007 #5


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    No, both derivatives should exist.

    [tex]lim _{h \to 0^-}\frac{f(h) - f(0)}{h}[/tex]

    [tex]=\lim _{h \to 0^+}\frac{f(-h) - f(0)}{-h}[/tex]

    [tex]= -\lim _{h \to 0^+}\frac{f(-h) - f(0)}{h}[/tex]

    [tex]= -\lim _{h \to 0^+}\frac{f(h) - f(0)}{h}[/tex]

    The left-sided derivative is just the negative of the right-sided derivative. If X exists, then so does -X.
  7. Mar 12, 2007 #6
    But the question says, it has only a one sided derivative. One of the derivatives does not exist.
  8. Mar 12, 2007 #7


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    AKG is right though.
  9. Mar 12, 2007 #8


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    That's not what the question says at all. Let's be clear. If f is a function, and x is some point in the domain, let L denote the left derivative of f at x, R the right derivative of f at x, and D the derivative of f at x. It's possible that some of these numbers don't exist, so I might right R = d.n.e. for example. Now if I write R = a, L = b, then understand that by using different symbols a and b, I mean a and b to be different, but I mean both of them to be real numbers, otherwise I'd write d.n.e. So we have the following possibilities:

    I) L = d.n.e., R = d.n.e., D = d.n.e.
    II) L = a, R = d.n.e., D = d.n.e.
    III) L = d.n.e., R = b, D = d.n.e.
    IV) L = c, R = d, D = d.n.e.
    V) L = e, R = e, D = e

    The questions asks to prove the existence of a one-sided derivative at 0, so it's asking you to prove that we're not in case I (i.e. we're in case II, III, IV, or V). It's not asking you to prove that it has only one one-sided derivative, i.e. it's not asking you to prove that you're either in case II or III. In this problem, you are in either case IV or case V. You're in case V iff any one of L, R, or D is 0.

    So when it asks: "Show that f has a finite one-sided derivative at 0" it means "Show that f has at leaste one finite one-sided derivative at 0", not "Show that f has only one finite one-sided derivative at 0." If I say "Show that 32 has an even divisor" I'm certainly not asking you to show that it has only one even divisor, I'm asking you to show there exists an even divisor, i.e. that there is at least one even divisor.
  10. Mar 13, 2007 #9
    Yeah. I made a mistake I guess. The RHD comes out to be +-1 and the LHD comes out to be -+1.
    [tex]lim_{h\to 0^+}=\frac{\sqrt(1-\sqrt(1-(0+h)^2)}{h}[/tex]
    which gives you [tex]+_-1[/tex] and similarly, the LHD comes out to be [tex]-_+1[/tex].
    Is this correct?
  11. Mar 13, 2007 #10


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    Uhm... What do you mean by [tex]+_-1[/tex], and [tex]-_+1[/tex]?!?! :confsed:
    You are correct up to this point. To continue, one can simply take the square root of h2, i.e:
    [tex]\lim_{h \rightarrow 0 ^ +} \frac{|h|}{h \sqrt{1 + \sqrt{1 - h ^ 2}}}[/tex]
    Since, h tends to 0+, i.e tends to 0 from the right, so that means h > 0, right?
    So |h| = h, the expression becomes:
    [tex]= \lim_{h \rightarrow 0 ^ +} \frac{h}{h \sqrt{1 + \sqrt{1 - h ^ 2}}} = \lim_{h \rightarrow 0 ^ +} \frac{1}{\sqrt{1 + \sqrt{1 - h ^ 2}}} = \frac{1}{\sqrt{2}}[/tex].
    Can you get this? :)
    Can you do the same to:
    [tex]= \lim_{h \rightarrow 0 ^ -} \frac{\sqrt{h ^ 2}}{h \sqrt{1 + \sqrt{1 - h ^ 2}}}[/tex]
    Last edited: Mar 13, 2007
  12. Mar 14, 2007 #11
    Yes. Looks like Im really on carless mistake making spree this week. Thank you.
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