Relative Minimums and Maximums — Question 7

Question 7 – Solution

Strategy. Match the quadratic terms with $\tfrac 12(h,k)H(h,k)^{T}$, then show that the leading signs dominate the smaller remainder.

Step 1: Hessian The second-order Taylor term is $$\frac 12\left(f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\right).$$ Matching coefficients gives $$f_{xx}=6,\qquad f_{xy}=4,\qquad f_{yy}=-2,$$ so $$\boxed{H(P)=\begin{pmatrix}6&4\\4&-2\end{pmatrix}}.$$

Step 2: Second derivative test $$D=(6)(-2)-4^2=-28<0.$$ Therefore $P$ is a saddle point.

Step 3: Remainder verification Along $k=0$, $$f(P+(h,0))-5=3h^2+o(h^2),$$ which is positive for all sufficiently small nonzero $h$. Along $h=0$, $$f(P+(0,k))-5=-k^2+o(k^2),$$ which is negative for all sufficiently small nonzero $k$. The little-$o$ terms are eventually smaller in magnitude than, for example, half of the displayed quadratic terms. Thus nearby values occur on both sides of $5$, rigorously confirming the saddle.