Answer
(a)
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{}\\
{{\bf{Point}}}&{{\bf{Type}}}\\
{\left( {0,3} \right)}&{{\rm{saddle{\ }point}}}\\
{\left( {1,3} \right)}&{{\rm{local{\ }minimum}}}\\
{\left( { - 1,3} \right)}&{{\rm{local{\ }minimum}}}
\end{array}$
(b) the minimum value of $f$ is $-10$.
Work Step by Step
(a) We have $f\left( {x,y} \right) = {x^4} - 2{x^2} + {y^2} - 6y$.
The partial derivatives are
${f_x} = 4{x^3} - 4x$, ${\ \ \ \ }$ ${f_y} = 2y - 6$
To find the critical points we solve the equations ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 4{x^3} - 4x = 0$
$x\left( {{x^2} - 1} \right) = 0$
$x = 0$, ${\ \ \ \ }$ $x = \pm 1$
${f_y} = 2y - 6 = 0$
$y = 3$
So, the critical points are $\left( {0,3} \right)$, $\left( {1,3} \right)$ and $\left( { - 1,3} \right)$.
The Second Derivative Test:
We evaluate the second partial derivatives:
${f_{xx}} = 12{x^2} - 4$, ${\ \ \ }$ ${f_{yy}} = 2$, ${\ \ \ }$ ${f_{xy}} = 0$
Using the Second Derivative Test, we determine the nature of the critical points and list the results in the following table:
$\begin{array}{*{20}{c}}
{{\bf{Critical}}}&{{f_{xx}}}&{{f_{yy}}}&{{f_{xy}}}&{{\bf{Discriminant}}}&{}\\
{{\bf{Point}}}&{12{x^2} - 4}&2&0&{D = {f_{xx}}{f_{yy}} - {f_{xy}}^2}&{{\bf{Type}}}\\
{\left( {0,3} \right)}&{ - 4}&2&0&{ - 8}&{{\rm{saddle{\ }point}}}\\
{\left( {1,3} \right)}&8&2&0&{16}&{{\rm{local{\ }minimum}}}\\
{\left( { - 1,3} \right)}&8&2&0&{16}&{{\rm{local{\ }minimum}}}
\end{array}$
(b) Notice that $f\left( {x,y} \right) = {x^4} - 2{x^2} + {y^2} - 6y$ can be written as
$f\left( {x,y} \right) = {\left( {{x^2} - 1} \right)^2} - 1 + {\left( {y - 3} \right)^2} - 9$
$ = {\left( {{x^2} - 1} \right)^2} + {\left( {y - 3} \right)^2} - 10$
Since ${\left( {{x^2} - 1} \right)^2} \ge 0$ and ${\left( {y - 3} \right)^2} \ge 0$, we conclude that the minimum value of $f$ is $-10$. From the results in part (a) we know that the minimum occurs at the critical points: $\left( {1,3} \right)$ and $\left( { - 1,3} \right)$.
