Question

Sketch the graph of an arbitrary function \(f\) satisfying the given conditions. State whether the function has any extrema or saddle points. (There are many correct answers.) $$ \begin{aligned} &f_{x}(2,1)=0, \quad f_{y}(2,1)=0\\\ &f_{x}(x, y)\left\\{\begin{array}{ll} >0, & x<2 \\ <0, & x>2 \end{array}, \quad f_{y}(x, y)\left\\{\begin{array}{ll} >0, & y<1 \\ <0, & y>1 \end{array}\right.\right.\\\ &f_{x x}(x, y)<0, f_{y y}(x, y)<0, \text { and } f_{x y}(x, y)=0 \text { for all }(x, y) \end{aligned} $$

Short answer

The function \(f\) has a saddle point at (2, 1). There are no other extrema or saddle points, based on the provided conditions. The function increases as we approach (2, 1) from left along x-axis and from below along y-axis, and decreases as we move away from (2, 1) along x-axis to the right and y-axis upwards.

Step-by-step solution

Identify the Critical Points

The first condition \(f_x(2,1) = 0\) and \(f_y(2,1) = 0\) suggest that we have a critical point at (2, 1). That is the point where both first order partial derivatives are zero.

Check Behavior near Critical Points

The next two conditions give information about the behavior of the function \(f\) near the critical point (2, 1). For \(f_x(x,y)\), if \(x < 2\), it is positive, indicating that \(f\) is increasing as \(x\) increases when \(x < 2\); if \(x > 2\), it is negative, indicating that \(f\) is decreasing as \(x\) increases when \(x>2\). On the other hand, for \(f_y(x,y)\), if \(y < 1\), it is positive indicating that \(f\) is increasing as \(y\) increases when \(y < 1\), if \(y > 1\), it is negative indicating that \(f\) is decreasing as \(y\) increases when \(y>1\).

Check the Nature of Critical Points

The last conditions state that the second derivatives \(f_{xx}(x,y)\) and \(f_{yy}(x,y)\) are negative, and \(f_{xy}(x,y) = 0\) for all \((x,y)\). Since \(f_{xx}<0\) and \(f_{yy}<0\) it suggests that the function \(f\) is concave downwards along both the x and y directions. The condition \(f_{xy}=0\) indicates there is no mixed partial derivative, suggesting that there is no twist in the graph of \(f\). Considering that \(f_x\), \(f_y\) changes sign across the critical point (2, 1), and on knowing the function is concave downwards, we can establish that (2, 1) is a saddle point.