Question

Derek found a function that approximately models the population of iguanas in a reptile garden, where x represents the number of years since the iguanas were introduced into the garden.
i(x) = 12(1.9)^x


Rewrite this function in a form that reveals the monthly growth rate of the population of iguanas in the garden. Round the growth factor to the nearest thousandth.

Answer 1

Answer:

$i(x)=12 \times (1+\frac{0.9}{12})^{12x}$ and growth rate factor is 0.075

Step-by-step explanation:

The function that models the population of iguanas in a reptile garden is given by $i(x)=12 \times (1.9)^{x}$, where x is the number of years.

Since, $i(x)=12 \times (1.9)^{x}$

i.e. $i(x)=12 \times (1+0.9)^{x}$.

Therefore, the monthly growth rate function becomes,

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{x \times 12}$.

i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$.

Hence, the monthly growth rate is i.e. $i(x)=12 \times (1+\frac{0.9}{12})^{12x}$.

Also, the growth factor is given by $\frac{0.9}{12}$ = 0.075.

Thus, the growth factor to nearest thousandth place is 0.075.