Find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point.
\(f(x, y) = x^2 + y^3 − 3y + 1\)
The partial derivative of f:
$$ \[\begin{multline*} \begin{split} f_x(x,y) = 2x \\ f_y(x,y) = 3y^2 - 3 \\ \end{split} \end{multline*}\] $$
Each partial derivative is always defined. Then to find the critical points we set each to 0 and solving for x and y, we find
\[ f_x(x,y) = 0 \space => \space x = 0 \\ f_y(x,y) = 0 \space => \space y = ± 1\\ D = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - f_{xy}^2(x_0,y_0)\\ \]
The 2nd derivatives of f are:
\[ \begin{multline*} \begin{split} f_{xx}(x_0,y_0) = 2\\ f_{yy}(x_0,y_0) = 6y_0\\ f_{xy}(x_0,y_0) = 0 \\ \end{split} \end{multline*} \]
As the 2 critical points (x0, y0) are (0,1) and (0,-1). We’ll check for D.
For 1st critical point (0,1),
D = 2 * 6*1 - 0 = 12
For 2nd critical point (0,-1),
D = 2 * 6*(-1) - 0 = -12
In conclusion, what we find is
there is a relative minimum for f at (0,1) as D > 0 and fxx(x0,y0) > 0 and,
there is a saddle point for f at (0,-1) as D < 0.