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

`i(x)=12 * (1+(0.9)/(12))^(12x)` and growth rate factor is 0.075

Explanation:

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

Since,
`i(x)=12 * (1.9)^(x)`

i.e.
`i(x)=12 * (1+0.9)^(x)`.

Therefore, the monthly growth rate function becomes,

i.e.
`i(x)=12 * (1+(0.9)/(12))^(x * 12)`.

i.e.
`i(x)=12 * (1+(0.9)/(12))^(12x)`.

Hence, the monthly growth rate is i.e.
`i(x)=12 * (1+(0.9)/(12))^(12x)`.

Also, the growth factor is given by
`(0.9)/(12)` = 0.075.

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