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Section 8.2 A Saddle Point Example

Example 8.9. Saddle Point.

Consider the function
\begin{equation} f(x,y) = \frac{y^2}{3} - \frac{x^2}{3}.\tag{8.2} \end{equation}
Show that \(f\) has a saddle point at \((0,0)\text{.}\)

Solution.

As we previously studied, this function represents a hyperbolic paraboloid, which is shown in Figureย 8.10. Note that the first partial derivatives are
\begin{equation} f_x = -\frac{2}{3}x \quad \text{and} \quad f_y = \frac{2}{3}y.\tag{8.3} \end{equation}
The partial derivatives exist everywhere, so the only point at which a local extremum can occur is the point \((0,0)\text{.}\) Note that along the \(x\)-axis the function takes negative values, \(f(x,0) = -x^2/3 \lt 0\text{,}\) whereas along the \(y\)-axis the function takes positive values, \(f(0,y) = y^2/3 \gt 0\text{,}\) as shown in Figureย 8.11. Hence inside every open disk around \((0,0)\text{,}\) there are points \((x,y)\) such that \(f(x,y) \gt f(0,0)\) and also there are points such that \(f(x,y) \lt f(0,0)\text{,}\) which means \((0,0)\) is a saddle point. Note that in this example the two partial derivatives are zero at \((0,0)\text{;}\) however, the function does not have a local extremum at this point.
Figure 8.10. The hyperbolic paraboloid \(z = y^2/3 - x^2/3\text{.}\) Along the \(x\)-axis the surface falls below the origin, and along the \(y\)-axis it rises above the origin, so \((0,0)\) is a saddle point.
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Figure 8.11. The traces of \(z = y^2/3 - x^2/3\) along the two coordinate planes. In the plane \(y = 0\) the trace \(z = -x^2/3\) opens downward, while in the plane \(x = 0\) the trace \(z = y^2/3\) opens upward, so \((0,0)\) is a saddle point.
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