Let x' be the unique solution of the linear system Ax=b. Find a differentiable function , so that x' is the only minimum.
I would say $f(x) = (x-x')^2$ but I never use the property of x' so that doesn't seem right. I was thinking of something like $f(x)=A^{-1}b$ but how do I know that A is invertible ?
If $x'$ is the unique solution of the linear system $Ax = b$, then $A$ is invertible.
A function such that $x'$ is the unique minimum build up using both $A$ and $b$ is:
$$f(x) = (Ax - b)^\top(Ax-b) = \\ =x^\top A^\top A x - 2b^\top Ax + b^\top b.$$
Differentiating, you get:
$$\frac{df}{dx} = 2A^\top A x - 2A^\top b = 2A^\top(Ax - b),$$
which is null for $x = x'$ (indeed $Ax'-b = 0$).
Notice also that the Hessian matrix is $$H = 2A^\top A,$$ which is positive definite, and hence all its eigenvalues are positive $\Rightarrow x'$ is the unique minimum.