Respuesta :
[tex]\begin{gathered} \text{Given} \\ S(t)=1000t^2+450t-1000 \end{gathered}[/tex]
Part A: average rate of increase from 2 to 3 days
Recall the average rate of change formula which is defined as
[tex]\begin{gathered} \text{If }f(x)\text{ is the function, then its average rate of change is} \\ \frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a} \\ \text{where} \\ a\text{ is the lower bound of the interval} \\ b\text{ is the upper bound of the interval} \end{gathered}[/tex]In this instance, the function is S(t) where the lower bound is a = 2, and the upper bound is b = 3.
The average, solve first for S(2) and S(3)
[tex]\begin{gathered} \text{If }t=2 \\ S(t)=1000t^2+450t-1000 \\ S(2)=1000(2)^2+450(2)-1000 \\ S(2)=1000(4)+900-1000 \\ S(2)=4000+900-1000 \\ S(2)=3900 \\ \\ \text{If }t=3 \\ S(t)=1000t^2+450t-1000 \\ S(3)=1000(3)^2+450(3)-1000 \\ S(3)=1000(9)+1350-1000 \\ S(3)=9000+1350-1000 \\ S(3)=9350 \end{gathered}[/tex]Next, solve for the rate of change over the interval [2,3] using the formula
[tex]\begin{gathered} \frac{\Delta y}{\Delta t}=\frac{S(3)-S(2)}{3-2} \\ \frac{\Delta y}{\Delta t}=\frac{9350-3900}{1} \\ \frac{\Delta y}{\Delta t}=5450 \end{gathered}[/tex]Therefore, the rate of increase of snails from 2 to 3 days is 5450.
Part B: rate of increase from 0 to 2
Using the same formula as above, find the S(0) and S(2)
[tex]\begin{gathered} \text{If }t=0 \\ S(t)=1000t^2+450t-1000 \\ S(0)=1000(0)^2+450(0)-1000 \\ S(0)=-1000 \\ \\ \text{If }t=2 \\ \text{As solved earlier, }S(2)=3900 \end{gathered}[/tex]Substituting with a = 0, and b = 2, we have the following
[tex]\begin{gathered} \frac{\Delta y}{\Delta t}=\frac{S(2)-S(0)}{2-0} \\ \frac{\Delta y}{\Delta t}=\frac{3900-(-1000)}{2} \\ \frac{\Delta y}{\Delta t}=\frac{3900+1000}{2} \\ \frac{\Delta y}{\Delta t}=\frac{4900}{2} \\ \frac{\Delta y}{\Delta t}=2450 \end{gathered}[/tex]Therefore, the rate of increase of snails after 2 days is 2450.
