Answer
Hence, It has been proved that $f$ has a local or absolute minimum value at each critical point and has an infinite number of critical points at $(x, \dfrac{1}{2} x)$.
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum and local maximum or, a saddle point.
We are given that $f(x,y)=x^2+4y^2+-4xy+2$
it yields, $f_x=2x-4y, f_{xx}=2$
and $f_y=8y-4x, f_{xx}=8$ , $f_{xy}=-4$
After equating all first and second partial derivative to $0$, we find that $x=y$ and $y=x$. This means that we get an infinite number of critical points.
For $(x,y)=(x, \dfrac{1}{2} x)$
$D(x, \dfrac{1}{2} x)=2 \gt 0$ and $f_{xx} \gt 0$; If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
Hence, It has been proved that $f$ has a local or absolute minimum value at each critical point and has an infinite number of critical points at $(x, \dfrac{1}{2} x)$.
