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A scalar valued function is defined as $f(x)=x^TAx+b^Tx+c$ , where A is a symmetric positive definite matrix with dimension $n*1$ ; b and x are vectors of dimension  $n*1$ .The minimum value of $f(x)$ will occur when x equals.

Answer: $-(\frac{A^{-1}b}{2})$

How to solve this?
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To find the minimum value differentiate the  f(x) , Use matrix calculus for differentiation (https://en.wikipedia.org/wiki/Matrix_calculus)

f'(x) =$\frac{\partial (x^{T} Ax+b^{T}+c) }{\partial x} = \frac{\partial x^{T} Ax }{\partial x} + \frac{\partial b^{T} x}{\partial x} +\frac{\partial c}{\partial x}$

$\frac{\partial  x^{T} Ax }{\partial x} =(A^{T}+A)x$    (using denominator layout)

$ \frac{\partial b^{T} x}{\partial x}= b$   (using denominator layout)

$ \frac{\partial c }{\partial x} = 0$ (as c is a constant)

Hence

$ \frac{\partial x^{T} Ax }{\partial x} + \frac{\partial b^{T} x}{\partial x} +\frac{\partial c}{\partial x}= (A^{T}+A)x+b=0$   (equating f'(x) with zero for minimum)

$ (A^{T}+A)x+b =2Ax +b$  ( A is a Symmetric matrix So $ A^{T}=A$ )

hence  $ x=-(A^{-1}b) /2$ 

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