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The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although equality of mixed partials reduces this to three. The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant $D$ that replaces $f\text{\u2033}\left({x}_{0}\right)$ in the second derivative test for a function of one variable.
Let $z=f\left(x,y\right)$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $\left({x}_{0},{y}_{0}\right).$ Suppose ${f}_{x}\left({x}_{0},{y}_{0}\right)=0$ and ${f}_{y}\left({x}_{0},{y}_{0}\right)=0.$ Define the quantity
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To apply the second derivative test, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.
Let $z=f\left(x,y\right)$ be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point $\left({x}_{0},{y}_{0}\right).$ To apply the second derivative test to find local extrema, use the following steps:
Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:
Therefore, $x=\mathrm{-1}$ or $x=3.$ Substituting these values into the equation $y=\frac{3-2x}{2}$ yields the critical points $\left(\mathrm{-1},\frac{5}{2}\right)$ and $\left(3,-\frac{3}{2}\right).$
Step 2 involves calculating the second partial derivatives of $g\text{:}$
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