Find a continuous function on the reals where $f(x) >0$ and $f'(x) < 0$ and $f''(x) < 0$ We need to find a function $f(x)$ where $f(x) >0 $and $f'(x) < 0$ and $f''(x) < 0$ where $f$ is continuous for all real numbers.
We have tried $ f(x) = \sqrt{-x}$ however this is not defined for $x>0$ and therefore is only continuous where $x<0$.
Is there even such a function? Because $f$ is positive but decreasing (...increasingly)
Any help is appreciated. 
Thanks.

Such function does not exist!
If not, there is such $f$ concave, so
$$f(x)\leq f'(0)x +f(0),  \qquad \forall x \in \Bbb R$$
Since $f'(0)\lt0$, so if $x \gt -\dfrac{f(0)}{f'(0)}$, then
$$f(x)\leq f'(0)x +f(0)\lt0$$
