# Global unique minimum $\frac{\sqrt{1+x^2}}{2}+\sqrt{4+(3-x)^2}$

Let $F(x)= \frac{\sqrt{1+x^2}}{2}+\sqrt{4+(3-x)^2}$, how can I prove that this function has an absolute minimum and it is located at a single point x, and satisfy $$\frac{x}{2\sqrt{1+x^2}}=\frac{3-x}{\sqrt{4+(3-x)^2}}$$

My attempt:

First the derivative is $-\frac{x}{2\sqrt{1+x^2}}+\frac{3-x}{\sqrt{4+(3-x)^2}}$ then if $f'(x)=0$ we have that $x=2$ is the root of $f'(x)$, now how can I conclude that is the unique point that satisfied this equation? Do I need to compute second derivative ?

Thanks for your help.

## 1 Answer

You do not need to compute the second derivative. Rather, use the first derivative test.

To apply the test, find $f'(a)$, such that $a<2$ and find $f'(b)$, such that $b<2$. If there is a sign change between the two test values, a maximum or minimum occurs at $x=2$, since the function changes from decreasing to increasing, or vice versa. If no sign change occurs, there is neither a maximum or minimum at $x=2$.