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I am working with this function:
$$f(x)=x_1^2+2x_2^2+3x_3^2$$
With the $min f(x)$ in $-10<=x_i<=10, i=1,2,3$.

I was given two points: $p_1=(1,1,1)$ and $p_2=(-1,2,1)$.

Using these points I have to find the supporting planes of $f$ for $p_1$ and $p_2$ and computing the upper and lower bounds for it.

I know that using this expression $f(p_1)+nabla f(p_1)(x-p_1)$ I got a supporting plane but I have to find the upper and lower bounds for $min f(x)$.
This would mean formulating a master problem but I am confused on how to compute the supporting planes and formulate the problem to solve it.

Please could you give some help?

Many thanks!

The normal vector of the supporting plane in the point $p=(p_1,p_2,p_3)$ is
$$nabla f(p) = {2p_1, 4p_2, 6p_3}.$$
Equation of the supporting plane for the point $p_1 = (1,1,1)$ is
$$f(p_1) + nabla f(p_1)cdot(x-p_1) = 0,$$
$$1^2 + 2cdot1^2 + 3cdot 1^2 + {2,4,6}cdot{x_1-1,x_2-1,x_3-1}=0,$$
or
$$f_1(x) = 2x_1+4x_2+6x_3-6=0.$$

Equation of the supporting plane for the point $p_2 = (-1,2,1)$ is
$$f(p_2) + nabla f(p_2)cdot(x-p_2) = 0,$$
$$(-1)^2 + 2cdot2^2 + 3cdot 1^2 + {-2,8,6}cdot{x_1+1,x_2-2,x_3-1}=0,$$
or
$$f_2(x) = -2x_1+8x_2+6x_3-12=0.$$

The least value of function achieves in the stationary points or in the edges of the area.

The stationary points of $f(x)$ can be found from the equation $nabla f(p_m) = 0,$ then
$$p_m = {0,0,0},quad f(p_m) = 0.$$

The edges of the area belongs to the planes $x_1 = pm 10,quad x_2=pm10, x_3=pm10.$

begin{vmatrix}
x_1 & x_2 & x_3 & f(x) & in\
-10 & [-10,10] & [-10,10] & 2x_2^2+3x_3^2+100 & [100,600]\
10 & [-10,10] & [-10,10] & 2x_2^2+3x_3^2+100 & [100,600]\
[-10,10] & -10 & [-10,10] & x_1^2+3x_3^2+200 & [200,600]\
[-10,10] & 10 & [-10,10] & x_1^2+3x_3^2+200 & [200,600]\
[-10,10] & [-10,10] & -10 & x_1^2+2x_2^2+300 & [300,600]\
[-10,10] & [-10,10] & 10 & x_1^2+2x_2^2+300 & [300,600]\
end{vmatrix}

The least value $color{brown}{f(p_m)=0}$ achives in the stationary point (minimum) $color{brown}{p_m={0,0,0}}.$

Function $f(x)$ is convex, so upper bound $color{brown}{sup(min f(x))=100}.$

Let us find the lower bounds for $min f(x).$

The least value of linear function in the rectangle area achives in the vertex of the area.

begin{vmatrix}
x_1 & x_2 & x_3 & f_1(x) & f_2(x)\
-10 & -10 & -10 & -126 & -132\
-10 & -10 & 10 & -6 & -12\
-10 & 10 & -10 & -46 & 28\
-10 & 10 & 10 & 74 & 148\
10 & -10 & -10 & -86 & -172\
10 & -10 & 10 & 34 & -52\
10 & 10 & -10 & -6 & -12\
10 & 10 & 10 & 114 & 108\
end{vmatrix}

Support plane $f_1(x)$ leads to lower bound $inf(min f(x)) = min f_1(x) = -126,$

Support plane $f_2(x)$ leads to lower bound $inf(min f(x)) = min f_1(x) = -172.$

The best estimation is $color{brown}{inf(min f(x)) = -126}.$