9.6 Low Points and High Points

We're now ready to look for low points and high points on the graph of a function \( z \depends (x, y) \). We'll say that a point \( p : \Point_{(x, y)} \) is a level point for the function if both partials are zero at \( p \).

\[ \begin{align} \overset{x}{\diff} z &= 0 & \overset{y}{\diff} z &= 0 \end{align} \]

Level points are points that have a horizontal tangent plane. Finding level points is the first step to finding low points and high points. Once we've found a level point, we'll need some criteria to decide whether it is a low point, a high point, or something else.

Important.

Suppose \( p : \Point_{(x, y)} \) is a level point. We may classify \( p \) by computing the saddle discriminant.

The level point is a low point if the graph is bowl-shaped,

\[ \begin{align} \discriminant z &\lt 0, \end{align} \]

and the bends are positive,

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\bend}} z &\gt 0 & &\andSpaced & \operatorname{\overset{\mathit{y}}{\bend}} z &\gt 0. \end{align} \]

The level point is a high point if the graph is bowl-shaped

\[ \begin{align} \discriminant z &\lt 0, \end{align} \]

and the bends are negative

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\bend}} z &\lt 0 & &\andSpaced & \operatorname{\overset{\mathit{y}}{\bend}} z &\lt 0. \end{align} \]

A level point with a saddle-shaped graph, \( \discriminant z \gt 0 \), is neither a low point nor a high point.

Let's see a couple examples!

Example.

Consider the function

\[ \begin{align} z &\depends (x, y) \\ z &= x^3 - y^2 - 3 x. \end{align} \]

Let's use our tools to look for low points and high points! We'll start by finding slopes

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\slope}} z &= 3 x^2 - 3 \\ \operatorname{\overset{\mathit{y}}{\slope}} z &= - 2 y \end{align} \]

so that we can find level points. Setting the \( x \)-slope equal to zero, we find

\[ \begin{align} &3 x^2 - 3 = 0 \\ &3 x^2 = 3 \\ &x^2 = 1 \\ &x = 1 \orSpaced x = -1. \end{align} \]

Setting the \( y \)-slope equal to zero gives

\[ \begin{align} &{-2} y = 0 \\ &y = 0. \end{align} \]

We conclude that there are only two level points on our graph, \( (1, 0) \) and \( (-1, 0) \). Let's classify these level points! To this end, we've computed the \( x \)-bend, the \( y \)-bend, the warp, and the saddle discriminant.

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\bend}} z &= 3 x & \operatorname{\overset{\mathit{y}}{\bend}} z &= -1 & \warp z &= 0 & \discriminant z &= 12 x \end{align} \]

At \( (1, 0) \).

At the point \( (1, 0) \), the discriminant is positive, \( \discriminant z = 12 \), and so we've found a saddle. This level point is neither a low point nor a high point.

At \( (-1, 0) \).

At the point \( (-1, 0) \), the discriminant is negative, \( \discriminant z = -12 \), so our graph is bowl-shaped here. Both bends are negative, so our level point \( (-1, 0) \) must be a high point!

In the previous example we found a saddle to the right of a high point. If we graph the function, we'll find the following surface.

If you'd like to get a better intuition for this surface, you can take some vertical slices through the level points.

Example.

Let's look for low points and high points on the graph of the function

\[ \begin{align} z &\depends (x, y) \\ z &= x^2 - 3 x y + y^2 + 3. \end{align} \]

We'll start by calculating slopes

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\slope}} z &= 2 x - 3 y & \operatorname{\overset{\mathit{y}}{\slope}} z &= 2 y - 3 x \end{align} \]

so that we can look for level points.

\[ \begin{align} 2 x - 3 y &= 0 \andSpaced 2 y - 3 x = 0 \end{align} \]

The only way to satisfy both equations is if both \( x \) and \( y \) are zero. So the origin \( (0, 0) \) is the only level point for our function. Let's compute bends, warp, and the discriminant to classify the graph's shape.

\[ \begin{align} \operatorname{\overset{\mathit{x}}{\bend}} z &= 1 & \operatorname{\overset{\mathit{y}}{\bend}} z &= 1 & \warp z &= -3 & \discriminant z &= 5 \end{align} \]

A positive discriminant tells us that we've found a saddle. Our level point \( (0, 0) \) is neither a low point nor a high point.

Looking at a contour map for the previous example, we see the saddle at the origin.

Notice, however, that we found both a positive \( x \)-bend and a positive \( y \)-bend. If we had just looked at bend, we might have been fooled into thinking that our level point is a low point! It is important to calculate the saddle discriminant: both bend and warp play a role in determining whether a graph is saddle-shaped.