1. ## Second derivative

I want to be clear about the concept of second derivative.

I would be able to tell whether a critical point is maximum or minimum from a second derivative. If second derivative is negative, the critical point is a maximum. If the second derivative is positive, the critical point is a minimum.

What if the sign of the second derivative depends on the value of x.
For example, I already know $x \in [0,\infty]$, and my second derivative is $x - 3$. So the second derivative is negative when x<3, positive when x>3 and 0 when x = 3. What does it tell me about the critical point? Please note I don't know what the critical point(s) are yet. Does this second derivative indicate more than one critical points?

2. ## Re: Second derivative

Originally Posted by avisccs
I want to be clear about the concept of second derivative.

I would be able to tell whether a critical point is maximum or minimum from a second derivative. If second derivative is negative, the critical point is a maximum. If the second derivative is positive, the critical point is a minimum.

What if the sign of the second derivative depends on the value of x.
For example, I already know $x \in [0,\infty]$, and my second derivative is $x - 3$. So the second derivative is negative when x<3, positive when x>3 and 0 when x = 3. What does it tell me about the critical point? Please note I don't know what the critical point(s) are yet. Does this second derivative indicate more than one critical points?
If you don't know what a critical point is then the answer to this question won't make sense either. See here under the heading: Critical point of a single variable function.

3. ## Re: Second derivative

Originally Posted by topsquark
If you don't know what a critical point is then the answer to this question won't make sense either.
If there is no critical point, there is no reason to calculate the second derivative. Is it correct?

My function is super-complex with 2 variables, I can't do it using algebra, but I can optimise it using Optim function in R. I'm trying to understand the result, whether it is a local or global maxima. I tried plotting, but still not sure. One can never plot from - infinity to infinity. So I now try to find any indication of problems with my result, starting with one variable and will go onto 2 variables. I know the derivative for 2 variable functions is different from single variable function.

4. ## Re: Second derivative

Originally Posted by avisccs
If there is no critical point, there is no reason to calculate the second derivative. Is it correct?

My function is super-complex with 2 variables, I can't do it using algebra, but I can optimise it using Optim function in R. I'm trying to understand the result, whether it is a local or global maxima. I tried plotting, but still not sure. One can never plot from - infinity to infinity. So I now try to find any indication of problems with my result, starting with one variable and will go onto 2 variables. I know the derivative for 2 variable functions is different from single variable function.
A critical point is one where the function does not exist or where the first derivative is 0. Note that a critical point does not need to be at a relative maximum or minimum of the function. As you said in your first post, the second derivative test will tell you if your critical point (assuming the first derivative is 0) is a relative maximum or minimum. A critical point where the first derivative and second derivative are both 0 is an inflection point.

Perhaps you should post your function...

-Dan

5. ## Re: Second derivative

Originally Posted by topsquark
Perhaps you should post your function...
My function with parameters x and y is
$q log\frac{x a (e^{-y b} -c)}{a-y} + (v-q)log (1-\frac{x a (e^{-y b} -c)}{a-y}) + \sum_{i=1}^n [r_i log(w_i(1-x e^{-y d}))- w_i(1-x e^{-y d})]$

Critical point exists, but I am wondering whether or not there are more than one.
The critical point I found from Optim function is a maxima, but I do not know whether there is another maxima at range which I did not plot.