Given the function $f : \mathbb{R}^3 \to \mathbb{R}$ defined by $f(x,\,y,\,z) = (x - 1)^4 - (x - y)^4 - (y - z)^4$ it presents a unique critical point of coordinates $(1,\,1,\,1)$.
Unfortunately, the hessian matrix at that point is null, so you need to make a local $f$ study. The study I find it really complicated. The only interesting thing that I noticed was that $ f (x, \, 1, \, 1) = 0 $: this is enough to say that $(1,\,1,\,1)$ is neither minimum nor maximum for $f$?
Thank you.
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$\begingroup$No.
We have $f(1,1,1)=0$, $f(1,1,z)=-(1-z)^4$ and $f(x,x,x)=(x-1)^4$.
This shows: in every neighborhood of $(1,1,1)$ the function f has values $> f(1,1,1)$ and values $< f(1,1,1)$.
Hence in $(1,1,1)$ the function $f$ has neither a minimum nor a maximum .
$\endgroup$ 0 $\begingroup$You've shown that $(1,1,1)$ can't be a strict local min or local max.
Noting that $f(x,x,x) = (x-1)^4$, we see that $f(1,1,1) = 0$ is not a maximum.
Noting that $f(1,1,z) = -(1-z)^4$, we see that $f(1,1,1) = 0$ is not a minimum.
In fact, one may show that the function is a saddle. define the function $$ F(x,y,z) = ((x-1)^2,(x-y)^2,(y-z)^2) $$ which has an inverse defined in a neighborhood of $(1,1,1)$. If we define $g(x,y,z) = x^2- y^2 - z^2$, then we see that $f = g \circ F$. After showing that $g$ has a saddle point at $(0,0,0)$, we may conclude that $f = g \circ F$ has a saddle point at the corresponding input, namely $(1,1,1)$.
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