Is $\sqrt{x}$ concave?

Question

I have function $f(x)= \sqrt(x)$. To check is it concave or convex i am checkin $f''(x). $ Which is $ -\frac{1}{4x^{\frac{3}{2}}} < 0$
So the $f(x)$ is concave. Is it correct ?

And is is the same with $f(x)=arctan(x)$, $f''(x)= -\frac{2x}{(x^{2}+1)^{2}}<0$. It is concave ?

Answer 1

$\sqrt{x}$ is indeed sure concave in its domain, i.e. $[0,+\infty)$, but $\arctan(x)$ in concave only for $x>0$ (the inteval where the last inequality you wrote holds), but for $x<0$ it's convex and in $x=0$ it has an inflection point.