Secondary derivative test: intuition and proof

Secondary derivative test: intuition and proof

This post is an annotation of Dennis Auroux's wondeful video. .

Starting with a simple example function $f(x, y) = ax^2 + bxy + cy^2$. We want to find out its local maximum and minimum (or whether they exist). After some algebra of completing the square we get:

\begin{align*} f(x, y) &= \frac{y^2}{4a} \left( 4a \left( u + \frac{b}{2a} \right)^2 +4ac - b^2 \right) \end{align*}

Now whether f(x,y) has a local extremum largely depends on whether $b^2 - 4ac<0$, if yes, then $ \frac{y^2}{4a} \left( 4a \left( u + \frac{b}{2a} \right)^2 +4ac - b^2 \right) $ is either always greater or less than 0, $f(x, y)$ is either bounded above or bounded below by the x-y plane.

For the general case, we use the second derivative test (google it). The idea is from Taylor's expansion:

\begin{align*} f(x, y) &= f(x_0, y_0) + f_x(x - x_0) + f_y(y - y_0) + (1/2)f_{xx}(x - x_0)^2 + f_{xy}(x - x_0)(y - y_0) + (1/2)f_{yy}(y - y_0)^2 \end{align*}

at the critical point, $f_x = f_y = 0$, so

\begin{align*} f(x, y) &= f(x_0, y_0) + (1/2)f_{xx}(x - x_0)^2 + f_{xy}(x - x_0)(y - y_0) + (1/2)f_{yy}(y - y_0)^2 \\ \Delta f &= (1/2)f_{xx}(x - x_0)^2 + f_{xy}(x - x_0)(y - y_0) + (1/2)f_{yy}(y - y_0)^2 \\ \end{align*}

Around a local extremum, $\Delta f$ should be either alway positive or always negative, now did you hear the bell ringing? Yes, you need the quadratic formula!