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Next: Directional derivative and Gradient Up: Applications of the derivative. Previous: Taylor's formula.

Extrema of a function and saddle points.

Definition 6.1   Let f be a function of two variables x and y defined on an open domain $\mathcal{D}$ in the plane. A point (x0,y0) is a critical point for f if f is differentiable at (x0,y0) and if fx(x0,y0) = fy(x0,y0)=0.

Example 6.2   Let f(x,y)=xy. Then fx=y and fy=x. We have:

\begin{displaymath}f_x=f_y=0 \Longleftrightarrow x=y=0.
\end{displaymath}

The function f has a single critical point, namely the origin.

Example 6.3   Let f(x,y)=x2y+x-y. Then fx=2xy+1 and fy=x2-1. We have:

\begin{displaymath}f_x=f_y=0 \Longleftrightarrow \begin{cases}2xy+1=0 \\ x^2-1 =0.
\end{cases}\end{displaymath}

This system of equations has two solutions, i.e. the function f has two critical points, namely (1,-1/2) and (-1, 1/2).

Theorem 6.4 (second derivative test)   Let f be a function of two variables x and y defined on an open domain $\mathcal{D}$ in the plane. Suppose that (x0,y0) is a critical point for f and that f has second order partial derivatives at ((x0,y0). Then:
1.
If fxx < 0 and fxxfyy-fxy2 >0 at ((x0,y0), then f has a maximum at ((x0,y0).
2.
If fxx > 0 and fxxfyy-fxy2 >0 at ((x0,y0), then f has a minimum at ((x0,y0).
3.
If fxxfyy-fxy2 < 0 at ((x0,y0), then f has a saddle point at ((x0,y0).

This theorem is proved by use of Taylor's formula (see section  1).

Example 6.5   Take f(xy)=xy, as in Example  6.2. We have: fxx= 0, fxy=1 and fyy=0. Thus fxxfyy-fxy2 = -1 <0 and the function f has a saddle point at the origin (see Figure  1(a)).


   
Figure 1: Surfaces with a saddle point.
\begin{figure}
\mbox{\subfigure[$z=x^2-y^2$ ]{\epsfig{file=HyperbolicParaboloid....
...ad
\subfigure[$z=x^2y^3$ ]{\epsfig{file=x2y3.eps,height=7cm} }
}\end{figure}

Example 6.6   Let f(x,y)=x2+y2. Then fx=2x and fy=2y and the function has a single critical point (0,0). Now fxx= 2, fxy=0 and fyy=2. The second derivative test (i.e. Thm  6.4) gives: fxxfyy-fxy2 =4 >0 and fxx=2 >0, i.e. the function f has a minimum at (0,0) (see Fig.  2).


  
Figure 2: The graph of a function with a minimum.
\begin{figure}
\mbox{\epsfig{file=Paraboloid3.eps,height=5cm} }
\end{figure}

Example 6.7   Let f(x,y)=x2y3. Then fx=2xy3 and fy=3x2y2. The function has a critical point at (0,0), but the second derivative test is unconclusive here, as fxx=y3, fxy=6xy2 and fyy=6x2y, and all three are equal to 0 at (0,0) (see Figure  1(b)). Only ``hand work'' can bring us a conclusion (which conclusion?).

Example 6.8   Let f(x,y)=x3+xy+y3. You can check that the function has a saddle point at the origin and a local maximum at (-1,-1), equal to -1. See Figure  3.
  
Figure 3: z=x3+xy+y3.
\begin{figure}
\mbox{\epsfig{file=saddlemax.eps,height=5cm} }
\end{figure}


next up previous contents
Next: Directional derivative and Gradient Up: Applications of the derivative. Previous: Taylor's formula.
Noah Dana-Picard
2001-05-30