Answer
(a) $S'(T)$ is the rate at which the oxygen solubility changes at the given temperature T. The units are $\frac{(mg/L)}{^{\circ}F}$
(b) $S'(16) \approx -0.2~\frac{(mg/L)}{^{\circ}F}$
When the temperature is $16^{\circ}F$, the oxygen solubility is changing at a rate of $-0.2~\frac{(mg/L)}{^{\circ}F}$
Work Step by Step
(a) $S'(T)$ is the rate at which the oxygen solubility changes at the given temperature T. The units are $\frac{(mg/L)}{^{\circ}F}$
(b) On the graph, we can see that the slope at $T = 16^{\circ}F$ is approximately equal to the average rate of change over the interval $[0, 40^{\circ}F]$
We can estimate $S'(16)$:
$S'(16) \approx \frac{15~mg/L-7~mg/L}{0^{\circ}F - 40^{\circ}F} = -0.2~\frac{(mg/L)}{^{\circ}F}$
When the temperature is $16^{\circ}F$, the oxygen solubility is changing at a rate of $-0.2~\frac{(mg/L)}{^{\circ}F}$