Yes. 

Indeed, $f(0+)=\infty=f(\infty-)$. So, the smooth function $f$ attains a minimum at some critical point of $f$ in $(0,\infty)$, that is, at a root $x\in(0,\infty)$ of $g$, where 
$g(x)$ is of the same sign as $f'(x)$ for $x>0$.  

The function $g$ is concave on $(0,a/2]$ and convex on $[a/2,\infty)$. Also, $g(0)=-1<0$, $g'(0)=b>0$, and $g(\infty-)=\infty$. So, either (i) $g$ has a unique root in $(0,\infty)$ or (ii) two roots $u,v$ such that $0<u<v<\infty$. 

We only have to consider case (ii), in which $g<0$ on $(0,u)$ and $g>0$ on $(u,v)$. So, $f'<0$ on $(0,u)$ and $f'>0$ on $(u,v)$. So, $f$ is decreasing on $(0,u)$ and increasing on $(u,v)$. So, the minimizer of $f$ on $(0,\infty)$ is the smallest root, $u$, of $g$ on $(0,\infty)$. $\quad\Box$