# Derivatives of an odd function

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

Thanks,
hk

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MalleusScientiarum
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.

HallsofIvy
Homework Helper
h_k331 said:
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).
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.
If f''(x) > 0 on (a,b), then f''(x) < 0 on (-b,-a).
Yes, this is true. If f(x) is odd, then f'(x) is even and f"(x) is odd again.
If lim (x->a) f(x) = inf, then lim (x->-a) f(x) = -inf
Yes. Also if lim(x->)f(x)= L, then lim(x->-a)f(x)= -L.
If x = a is a vertical asymptote of f, then x = -a is also a vertical asymptote of f.
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".
If lim (x->inf) f(x) = L, then lim (x->-inf) f(x) = -L
Yes. In fact, as above, this is true for (x-> any number) as well.
If y = L is a horizontal asymptote of f, then y = -L is also a horizontal asymptote of f.
Yes. This is just a restatement of "If lim(x->inf)= L, then lim(x->-inf)(f)= -L
If f is odd and f is continuous on (-a,a), then f(0) = 0
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.

Thanks,
hk

Thank you for your reply HallsofIvy, I have a better understanding now.

hk