Consider the following problem: \begin{equation*} \begin{aligned} & \underset{x \in\mathbb{R}}{\text{minimize}} & & f(x) = \frac{1}{2}x^TAx - bx \\ \end{aligned} \end{equation*} where $b \in \mathbb{R^n} $ is a given vector and $A$ is $n$ x $n$ symmetric positive definite matrix, i.e., $x^TAx > 0$ for any $x$.
a) Let $x = [x_1, x_2, \dots, x_n]^T$. Write the function $f$ in the expanded form, as a function of $x_1, x_2, \dots , x_n$.
b) Find the critical points of $f$.
c) Determine if the critical points are solutions of the above problem.
For a) I expanded it in the following way: \begin{equation} f(x) = \frac{1}{2}(a_{11}x_1^2 +a_{12}x_1x_2 + \dots + a_{1n}x_1x_n+a_{21}x_1x_2+a_{22}x_2^2+ \dots + a_{2n}x_2x_n+ \dots +a_{n1}x_1x_n+a_{n2}x_2x_n+\dots+a_{nn}x_n^2) - (b_1x_1 + \dots +b_nx_n)\end{equation}
For b) I did the following: \begin{equation}f'(x) = Ax - b = 0 => x =A^{-1}b,\end{equation} so these vectors are the critical points.
And, finally for c) as $f''(x) = A$, which is the Hessian and as it is positive definite, then the critical points are the $min$ points and they are a solution to the problem. I'd like to know if I solved them correctly. Any advice would be very appreciated.