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I have the function $f:\mathbb{R}^3 \mapsto \mathbb{R}$ defined as $f(x, y, z) = xy+xz+yz+z-x$. I've calculated the Jacobian:

$$ J_f = (y+z-1, x+z, x+y+1) $$

Which, by setting $J_f = \vec{0}$, reveals that I have a stationary point at (-1, 0, 1). To find if it's a maxima or minima (or a saddle point) I derived the Hessian matrix. The Hessian matrix is a 3x3 symmetric matrix with zeroes in the main diagonal and ones in the other entries.

Now, $det(H_f) = 0(0-1) - 1(0-1) + 1(1-0) = 2 > 0$, hence this critical point should be a local maximum or minimum according to the second derivative test. However, the solution's manual sets this critical point as a saddle. Which one is correct?

The eigenvalues give me $\lambda_1 = 2, \lambda_2 = \lambda_3 = -1$.

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    $\begingroup$ I think we must read the statement of the second derivative test more carefully. $\endgroup$
    – littleO
    Apr 19 at 10:45

1 Answer 1

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The second derivative test (and the Hessian determinant) only works for bivariate functions.

For functions of three or more variables, one needs to use the eigenvalues. In this case, as we have both positive and negative eigenvalues, $(-1,0,1)$ must be a saddle point.

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