# Is $\sqrt{x}$ concave?

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 ?

$\sqrt{x}$ 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.
• I i will have $f(x)=arctan(x)$ and i will add that $x\geq 0$ then i can say that $f(x)$ is concave. – Mango Jun 15 '14 at 13:34
• It's the opposite: in order to have $\arctan(x)$ concave you have to impose $x>0$. – Dario Jun 15 '14 at 13:36
• what if i will have $x\geq 0$ then it is not concave. – Mango Jun 15 '14 at 13:40
• It depends on your definition of concave: there are the notion of "concave" and "strictly concave". In $x\geq0$ $\arctan(x)$ is concave, but not strictly concave. (The difference between the two notions translate in terms of the second derivative as the two conditions $f''\leq0$ or $f''<0$) – Dario Jun 15 '14 at 13:47