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(a) \(A = A^T\) ( \(\because\) A is a symmetric matrix)
\(f(x) = {1 \over 2} x^TAX + b^Tx\)
\(= {1 \over 2}Ax^2 + b^Tx\)
\(\bigtriangledown f(x) = f(x) = Ax + b ({\partial b^T x \over \partial x} = {\partial x^T b \over \partial x} = b)\)
\(\therefore \bigtriangle f(x) = Ax + baaa\)