Local minimum of the function:

Question

Find the local minimum of the function: $$\def\f{f(x_1,x_2,x_3)}\def\1{x_1}\def\2{x_2}\def\3{x_3}\def\n{\nabla}$$ $$\f=\1^2-2\1\2+2\2^2+\3^2 \text{ in } \mathbb{R}^3$$

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$\n\f=(2\1-2\2,-2\1+4\2,2\3) = \vec{0}$ Hence $\3=0$, and $\1=\2$, but $2\1=4\2$, hence $\1=\2=\3=0$

And the critical point is $(0,0,0)$ at $\f$ is $f(0,0,0)=0$.

Taking the Hessian matrix I can see that $\det(H(\1,\2,\3))=8$ and $f_{XX}=2$. Since these are both $\gt0$, this is a minimum.

Is this correct?

Answer 1

$$f(x_1 , x_2 , x_3 ) =(x_1 -x_2 )^2 +x_2^2 +x_3^2 $$ hence $$\min_{(x_1 ,x_2 ,x_3 )\in\mathbb{R}^3 } f(x_1 , x_2 , x_3 ) =f(0,0,0) =0$$