Answer
a) $P'(5)$ is the instantaneous rate of growth of yeast cells at $t=5$ hours. Its unit is cell/hour.
b) $P'(3)$ is larger than $P'(2)$.
c) The instantaneous rate of growth at $t=5$ is $51.72$ cells/hour.
Work Step by Step
a) $P'(5)$ is the derivative of $P$ at $t=5$, which is also the instantaneous rate of change of $P$ with respect to $t$ at $t=5$. Applying to the growth of yeast cells here, it is the instantaneous rate of growth of yeast cells at $t=5$ hours.
Its unit is cell/hour.
b) Since $P'(x_0)$ can also represent the slope of the tangent line to the curve at $x=x_0$, and the value of a slope would be translated into the direction and steepness of a line to the $x$-axis, to compare $P'(2)$ and $P'(3)$, we look at the curve at those points.
We see that at $x=2$ and $x=3$, the tangent lines at these points make acute angles with the $x$-axis, so both slopes of these lines are positive.
Next, we look at how steep the curve is at these two points. Basically, the steeper the curve is, the steeper its tangent line will be, and thus the larger the value of the slope there. Here, the curve at $x=3$ is steeper than that at $x=2$, so the slope of the tangent line at $x=3$ is larger than that at $x=2$.
This means $P'(3)$ is larger than $P'(2)$.
c) The instantaneous rate of growth at $t=5$, or $P'(5)$, is calculated by $$P'(5)=\lim_{h\to0}\frac{P(h+5)-P(5)}{h}$$ $$P'(5)=\lim_{h\to0}\frac{\Big(6.10(h+5)^2-9.28(h+5)+16.43\Big)-\Big(6.10\times5^2-9.28\times5+16.43\Big)}{h}$$ $$P'(5)=\lim_{h\to0}\frac{(6.10(h^2+10h+25)-9.28h-46.4+16.43)-122.53}{h}$$ $$P'(5)=\lim_{h\to0}\frac{6.10h^2+61h+152.5-9.28h-46.4+16.43-122.53}{h}$$ $$P'(5)=\lim_{h\to0}\frac{6.10h^2+51.72h}{h}=\lim_{h\to0}(6.10h+51.72)$$ $$P'(5)=6.10\times0+51.72=51.72(cells/hour)$$
