How to show that p=(0,0,0) is a stationary point of f=exp(x²+y²+z²)+4xy+2xz+2yz

Determine whether the stationary point p corresponds to local minimum, local maximum or saddle point.

Thanks for the help!!

First, compute the first partial derivatives.
f_x = 2x e^(x²+y²+z²) + 4y + 2z
f_y = 2y e^(x²+y²+z²) + 4x + 2z
f_z = 2z e^(x²+y²+z²) + 2x + 2y,

Since f_x(0,0,0) = f_y(0,0,0) = f_z(0,0,0) = 0, we conclude that (0,0,0) is a stationary point.
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To classify this, we apply the second derivative test.

f_xx = 2e^(x²+y²+z²) + 4x² e^(x²+y²+z²) ==> f_xx (0,0,0) = 2
f_xy = 4xy e^(x²+y²+z²) + 4 ==> f_xy (0,0,0) = 4
f_xz = 4xz e^(x²+y²+z²) + 2 ==> f_xz (0,0,0) = 2

f_yy = 2e^(x²+y²+z²) + 4y² e^(x²+y²+z²) ==> f_yy (0,0,0) = 2
f_yz = 4yz e^(x²+y²+z²) + 2 ==> f_yz (0,0,0) = 2

f_zz = 2e^(x²+y²+z²) + 4z² e^(x²+y²+z²) ==> f_zz (0,0,0) = 2
(Note: Equality of mixed partial derivatives certainly applies here!)

So, the Hessian (second partial derivatives) matrix at (0,0,0) equals
[2 4 2]
[4 2 2]
[2 2 2].

Its determinant D₃ equals 2 * 0 - 4 * 4 + 2 * 4 = -8 < 0.
The upper left 2 x 2 minor D₂ equals -12 < 0.
The upper left 1 x 1 minor D₁ equals 2 > 0.

Since these determinants are not all positive (or alternating starting with D₁ < 0, D₂ > 0, etc.),
we have a saddle point at (0,0,0).

I hope this helps!