## Exam 1

1. #### Question

Compute the Hessian of the function
 $\begin{array}{c}\multicolumn{1}{c}{f\left({x}_{1},{x}_{2}\right)=-7{x}_{1}^{2}-5{x}_{1}{x}_{2}-4{x}_{2}^{2}}\end{array}$

at $\left({x}_{1},{x}_{2}\right)=\left(-2,2\right)$. What is the value of the upper left element?

1. $-8$
2. $-5$
3. $-16$
4. $-14$
5. $-4$

#### Solution

The first-order partial derivatives are
 $\begin{array}{ccc}\multicolumn{1}{c}{f{\text{'}}_{1}\left({x}_{1},{x}_{2}\right)}& =\hfill & -14{x}_{1}-5{x}_{2}\hfill \\ \multicolumn{1}{c}{f{\text{'}}_{2}\left({x}_{1},{x}_{2}\right)}& =\hfill & -5{x}_{1}-8{x}_{2}\hfill \end{array}$

and the second-order partial derivatives are
 $\begin{array}{ccc}\multicolumn{1}{c}{f"{}_{11}\left({x}_{1},{x}_{2}\right)}& =\hfill & -14\hfill \\ \multicolumn{1}{c}{f"{}_{12}\left({x}_{1},{x}_{2}\right)}& =\hfill & -5\hfill \\ \multicolumn{1}{c}{f"{}_{21}\left({x}_{1},{x}_{2}\right)}& =\hfill & -5\hfill \\ \multicolumn{1}{c}{f"{}_{22}\left({x}_{1},{x}_{2}\right)}& =\hfill & -8\hfill \end{array}$

Therefore the Hessian is
 $\begin{array}{c}\multicolumn{1}{c}{f"\left({x}_{1},{x}_{2}\right)=\left(\begin{array}{cc}\hfill -14& \hfill -5\\ \hfill -5& \hfill -8\end{array}\right)}\end{array}$

independent of ${x}_{1}$ and ${x}_{2}$. Thus, the upper left element is: $f"{}_{11}\left(-2,2\right)=-14$.

1. False
2. False
3. False
4. True
5. False