What Does Changing Sign in Derivatives Mean for Completely Monotone Functions?

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Homework Statement



This is ralated to completely monotone functions. if a function and its successive derivative changes sign then what does this mean?

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The Attempt at a Solution

 
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Changes sign where? If a function, f, is monotone increasing then its derivative is always positive, but f may be positive for some values of x and negative for others. the second derivative then may be positive for some values of x and negative for others.
 
changes sign at every point on the domain.
if we have a function f(x)=exp(-a*x) then f'(x)=-a*exp(-a*x),
f''(x)=a^2*exp(-a*x) and so on...
similarly if we have f(x)=(a^2+x)^(-1/2) then f'(x)=(-1/2)*(a^2+x)^(-3/2)
and f''(x)(-1/2)*(-3/2)*(a^2+x)^(-5/2) and so on...
i think that if we have such kind of functions then there will be successively concave up( the slope will become steeper) and concave down( the slope will be come shallower). i want such kind of explanation.
 
As you said,
[tex]f=e^{-ax}[/tex]
[tex]f'=-ae^{-ax}[/tex] This does not change signs(a is const),its monotonous.
 
Last edited:
I know that such kind of functions are completely monotone functions. but its successive derivative changes sign. and you pointed out that this does not change sign, how? the function is positive and its derivative is negative and the second derivative will agian be positive.
 
aquarian11 said:
you pointed out that this does not change sign, how?
Read again, I said [itex]\ f'[/itex] is not changing signs.

aquarian11 said:
the function is positive and its derivative is negative and the second derivative will agian be positive.

[itex]f'' \mbox{and} f' \mbox{are both having same sign for}\ a<0[/itex]