Min value of $F(x) = x - \sqrt{1-x^2} $ Min value of $F(x) = x - \sqrt{1-x^2} $

But using derivatives  $F(x) = x - \sqrt{1-x^2} $

i stucked at $2\sqrt{1-x^2} + 2x = 0$

$\sqrt{1-x^2} &gt; 0 $

I get $-1&lt;x&lt;1$

Please help me?
 is there another way to solve it btw?
I think there may be something wrong in this (that means, your derivative): $2\sqrt{1-x^2} + 2x = 0$

My answer:


  First let's notice that the function domain is $[-1,1]$ for that $1-x^2\geqslant 0$. Then it is easy to see $f'(x)=0$ when $f(x)$ reaches to minimum. So that
  $$
f'(x)=\frac{x}{\sqrt{1-x^2}}+1
$$
  set $f'(x)=0$ we get $x=-\frac{1}{\sqrt{2}}$, and $f(x)=-\sqrt{2}$ where it is the minimum. Rigorously speaking, you need to check the derivative on both side to judge whether it is the min value, but in this question it is obvious.
Rearrange  

$$2\sqrt{1-x^2} + 2x = 0$$

to get

$$\sqrt{1-x^2} = -x$$

Square both sides

$$1-x^2=x^2$$

and solve for $x$.