# investigating the second derivative test

• Nov 12th 2008, 08:56 AM
chrisc
investigating the second derivative test
This is a question that is supposed to help my understanding of the second derivative test, but Im not sure how it does, or how to do this.

f(x,y) = a(x^2) + b(xy) + c(y^2)

by completing the square, show that if a/=0 then,

f(x,y) = a(x^2) + b(xy) + c(Y^2) = a[(x + yb/2a)^2 + (y^2)(4ac-b^2)/(4a^2)]
• Nov 12th 2008, 12:02 PM
Opalg
If you apply the first derivative test (putting partial derivatives equal to 0), you see that the only critical point of f(x,y)occurs at the origin. If you apply the second derivative test (calculate the second partial derivatives and look at $\tfrac{\partial^2f}{\partial x^2}\tfrac{\partial^2f}{\partial y^2} - \bigl(\tfrac{\partial^2f}{\partial x\partial y}\bigl)^2$), it tells you that there is a local max/min at the origin (depending on whether $\tfrac{\partial^2f}{\partial x^2}$ is negative or positive) if that expression is positive, and that there is a saddle point if it is negative.

On the other hand, if you look at the formula obtained by completing the square, you get exactly the same information: if $4ac-b^2<0$ then one of the squares is positive and the other one is negative, so there is a saddle point at the origin. If $4ac-b^2>0$ then both squares are positive, and the function will have a local max/min at the origin depending on whether a is negative or positive.

So it looks as though the aim of this exercise is to persuade you that the second derivative test works, by seeing that it gives the right answer in this case.