It's all in the pdf.
10x in advance.
If the second derivative is positive at a critical value then the critical value is a relative minimum. If the second derivative is negative then the critical value is a relative maximum.
-Dan
EDIT: Sorry, I hadn't looked at the actual problem. The test I'm referring to is for one dimensional Calculus. I'll let someone who knows the subject better take this one.
Ok, I couldn't open your PDF and your image won't appear either.
But here is what I think you are asking
Let denote a critical point of
In other words
or is undefined
Now define
Now if then the point is a relative min.
If then is a relative max.
If then is a Saddle Point of
And if no conclusion can be drawn.
Is this what you are asking about? If not just respond saying so.
Mathstud.
I rescan everything in BMP so I hope you can see it now.
What I'm talking about is a function with a condition, after I find all the points, how do I determinate which one is maximum and which one is minimum, I know I need to use d^2f somehow but I dont know how.
10x in advance.
Im really sorry, but once again I cannot see it. If you are not talking about finding relative extrema then I assume this is a question of either Absolute Extrema in a bounded region or a question of Lagrange Multipliers.
Now this may sound novel, but have you tried just evaluating all the points, and from there since you know that they are all relative extrema you can seee which are the absolute max and mins?
You right, I'm talking about Lagrange Multipliers, lets say I have a function f(x,y) and also a condition g(x,y), and I want to build a function l(x,y) = f(x,y) + gama * g(x,y). From here I find all the points, my question is how can I tell which one is minimum and which one is maximum? can I just put them in f(x,y) and evaluate it from there? shouldn't I use d^2l or something like that?
10x again.
Ok, so I finally see your page it says
Maximize
Under the condition that
So defining
We need to solve the following system of equations
So computing
We see that we must solve
So we see that the four solutions are
and
and
and
So testing we get
You can draw your conclusions from there.