I have a question for
$$f(x) = x+ \sqrt{1-x^2}$$
It is clear by inspection that $x= -1$ is the minimum, because for the square root. If we have $x$ smaller then we have an imaginary number, if x is larger, then the second term can be only positive and add to the total value. My question is that we generally have an optimality condition that says $f'(x^*) = 0$, for continuously differentiable. But
$$f'(x) = 1 - \frac{x}{\sqrt{1-x^2}}$$
and if we say $x \rightarrow -1$ then we get $$\lim_{x \rightarrow -1} f'(x)= 1 - \frac{x}{\sqrt{1-x^2}}$$ which gives us undefined. So the reason the optimality condition does not hold is because we are not differentiable at $-1$?
Since your domain is the closed interval [-1,1], you need to check the value of your function at the endpoints as well as checking for local minimizers. You do not need to check derivatives at endpoints.