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Derivatives of an odd function

  1. Jul 9, 2005 #1
    I've come to a few assumtions that my text hasn't covered yet, and I was hoping someone could confirm these for me or let me know where I might be missing something.

    Assuming f is an odd function and a > 0 and b > 0 and a < b:

    If f'(x) > 0 on (a,b), then f'(x) < 0 on (-b,-a).

    If f''(x) > 0 on (a,b), then f''(x) < 0 on (-b,-a).

    If lim (x->a) f(x) = inf, then lim (x->-a) f(x) = -inf

    If x = a is a vertical asymptote of f, then x = -a is also a vertical asymptote of f.

    If lim (x->inf) f(x) = L, then lim (x->-inf) f(x) = -L

    If y = L is a horizontal asymptote of f, then y = -L is also a horizontal asymptote of f.

    If f is odd and f is continuous on (-a,a), then f(0) = 0

  2. jcsd
  3. Jul 9, 2005 #2
    The derivative of an odd function is an even function. Anything else you want to know can be taken from that.

    Also, to be "odd" requires that f(0) = 0. Figure out why.
  4. Jul 10, 2005 #3


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    No! If f(x) is odd, then f'(x) is even. f' will have the same sign (indeed exactly the same values) for both (a,b) and (-b,-a).
    An easy example: if f(x)= x, then f'(x)= 1 for all x.
    Yes, this is true. If f(x) is odd, then f'(x) is even and f"(x) is odd again.
    Yes. Also if lim(x->)f(x)= L, then lim(x->-a)f(x)= -L.
    Yes. This is essentially just a restatement of the previous statement of "If lm(x->a) f(x)= inf, then lim(x->-a)f(x)= -inf".
    Yes. In fact, as above, this is true for (x-> any number) as well.
    Yes. This is just a restatement of "If lim(x->inf)= L, then lim(x->-inf)(f)= -L
    You don't need "and f is continuous on (-a,a)". If f is odd then f(-x)= -f(x) for all x. Taking x= 0, f(-0)= f(0)= -f(0) so 2f(0)= 0 and f(0)= 0.

  5. Jul 10, 2005 #4
    Thank you for your reply HallsofIvy, I have a better understanding now.

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