Note that the equation given for $\dfrac{dv}{dt}$ is a scalar equation, so $v$ actually represents speed rather than velocity.

Speed is bounded strictly between $0 \le v \le v_T = \dfrac{mg}{k}$, so the signs of $\dfrac{dv}{dt}$ and $\dfrac{d^2v}{dt^2}$ need only be calculated between those two bounds ...

for $v = \dfrac{mg}{2k}$ ...

$\dfrac{dv}{dt} = g - \dfrac{k}{m} \cdot \dfrac{mg}{2k} = g - \dfrac{g}{2} = \dfrac{g}{2} > 0$

$\dfrac{d^2v}{dt^2} = -\dfrac{k}{m} \cdot \dfrac{dv}{dt} = -\dfrac{k}{m} \cdot \dfrac{g}{2} < 0$

so, the graph of speed, $v$, versus time is increasing and concave down.