# Extrema points, function of three variable

#### GreenGoblin

##### Member
"Show that if a>b>c>0, then the function $f(x,y,z) = (ax^{2} + by^{2} + cz^{2})e^{-x^{2}-y^{2}-z^{2}}$ has two local maxima, one local minima, and four saddle points"

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#### Jameson

Staff member
Your Latex isn't rendering and has tons of unnecessary code. Can you clean it up to make it more readable please?

#### GreenGoblin

##### Member
I have edited it out from hand and started again
Thanks

What I have done is taken derivatives of the partials in the first and second order, do i need to do third order (fxxx, fxxy, fxxz, fxyy etc?) i know the second derivative test for function of two variables, but for three variables? do ineed third derivative test? how does it work, is there another test? what determinants do i need to take

#### Alexmahone

##### Active member
What I have done is taken derivatives of the partials in the first and second order, do i need to do third order (fxxx, fxxy, fxxz, fxyy etc?) i know the second derivative test for function of two variables, but for three variables? do ineed third derivative test? how does it work, is there another test? what determinants do i need to take
First, find the points where $\displaystyle f_x,\ f_y,\ f_z$ vanish. These are the critical points. To find the nature of the critical points, use the method explained in this paper.

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#### HallsofIvy

##### Well-known member
MHB Math Helper
In the strictest sense, the "derivative" of a function of three variables, f(x,y,z), at $$(x_0, y_0, z_0)$$, is the linear function, from $$R^3$$ to R, $$U(x,y,z)= f_{x}(x_0,y_0,z_0)(x- x_0)+ f_y(x_0, y_0, z_0)(y- y_0)+ f_z(x_0,y_0,z_0)(z- z_0)$$ which we can think of as the vector dot product $$<f_x(x_0, y_0,z_0), f_y(x_0,y_0,z_0), f_z(x_0, y_0, z_0)>\cdot<(x- x_0, y-y_0, z-z_0>$$ and so can be "represented" by the gradient vector $$<f_x(x_0, y_0,z_0), f_y(x_0, y_0, z_0), f_z(x_0, y_0, z_0)>$$.

In that same sense, the second derivative of a function of three variables is the linear function from $$R^3$$ to the set of such gradient vectors which are themselves in $$R^3$$ which can be represented by the three by three matrix:
$$\begin{bmatrix}f_{xx} & f_{xy} & f_{xz} \\ f_{yx} & f_{yy} & f_{yz} \\ f_{zx} & f_{zy} & f_{zz}\end{bmatrix}$$

Now, because the mixed derivatives are equal: $$f_{xy}= f_{yx}$$, $$f_{yz}= f_{zy}$$, and $$f_{xz}= f_{zx}$$, that is a symmetric matrix which means it is diagonalizable. That is, there exist some coordinates system, x', y', z', in which all mixed derivatives are 0 and the matrix is
$$\begin{bmatrix}f_{x'x'} & 0 & 0 \\ 0 & f_{y'y'} & 0 \\ 0 & 0 & f_{z'z'}\end{bmatrix}$$
where those second derivatives, $$f_{x'x'}$$, $$f_{y'y'}$$, $$f_{z'z'}$$, evaluated at $$(x_0, y_0, z_0)$$ are the eigenvalues of the matrix.

And that, in turn, means that in that coordinate system we can write
$$f(x',y',z')= f(x_0,y_0, z_0)+ f_{x'x"}(x'- x_0)^2+ f_{y'y'}(y'- y_0)^2+ f_{z'z'}(z'- z_0)^2$$

Now we can see: if all of those eigenvalues are positive, [itex](x-0, y_0, z_0)[/tex] is a minimum, if all negative, a minimum, if some positive and some negative then a saddle point. Now, if this were two variables, x and y, say, our matrix would be 2 by 2:
$$\begin{bmatrix}f_{xx} & f_{xy} \\ f_{yx} & f_{zz}\end{bmatrix}$$
or, in the x', y', z' coordinate system in which it is diagonal,
$$\begin{bmatrix}f_{x'x'} & 0 \\ 0 & f_{y'y'}\end{bmatrix}$$
Notice that that last matrix has determinant $$f_{x'x'}f_{y'y'}$$ and so is positive if and only if $$f_{x'x'}$$ and $$f_{y'y'}$$ have the same sign and negative if and only if they have different sign. But the determinant is independent of the coordinate system so we can say that f has a saddle point if and only if $$f_{xx}f_{yy}- f_{xy}^2< 0$$ and a max or min if it is positive (in that case $$f_{xx}$$ and $$f_{yy}$$ must have the same sign so you can check either to see whether it is a max or min).

Unfortunately, it isn't that easy with three variables. If the determinant is positive, it might be that all three eigenvalues are positive (so a minimum) or that one is positive and the other two negative (a saddle point) or if the determinant is negative, it might be that all three eigenvaues are negative (so a maximum) or that one is negative and the other two positive (a saddle point). You really need to identify all three eigenvalues of the "second derivative matrix" in order to know what you have.