If $z=\begin{bmatrix}x & x^2\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}\begin{bmatrix}x\\x^2\end{bmatrix}$, then $\begin{pmatrix}\frac{d^2z}{dx^2}\end{pmatrix}_{x=-1}$ is:
Answer & explanation
Correct answer: option 2
$z=\begin{bmatrix}x & x^2\end{bmatrix}\begin{bmatrix}1 & -1\\2 & 3\end{bmatrix}\begin{bmatrix}x\\x^2\end{bmatrix}=\begin{bmatrix}x+2x^2 & -x+3x^2\end{bmatrix}\begin{bmatrix}x\\x^2\end{bmatrix}$
$Z=x^2+2x^3-x^3+3x^4$ $⇒Z=x^2+x^3+3x^4$
$\frac{dz}{dx}=2x+3x^2+12x^3$ $⇒ \frac{d^2z}{dx^2}=2+6x+36x^2$
$\frac{d^2z}{dx^2}|_{x=-1}=2-6+36(-1)^2=2-6+36=32$