Question

ednas grandparents put money in a savings account when she was born. the function $f(x)=6000cdot1.08^{x}$ represents the value of the account when edna is $x$ years old. edna compares the average rate of change in the first two years after she was born to the two years before she turned 18. complete the paragraph. between years 0 and 2, the account increased by dollars per year. between years 16 and 18, the account increased by dollars per year. the account increased by amount in the two years before edna turned 18.

Explanation:

Step1: Calculate the average rate of change between years 0 and 2

The average rate of change formula is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b=2$, $f(x)=6000\times1.08^{x}$.

First, find $f(0)$: $f(0)=6000\times1.08^{0}=6000$.

Then, find $f(2)$: $f(2)=6000\times1.08^{2}=6000\times1.1664 = 6998.4$.

The average rate of change is $\frac{f(2)-f(0)}{2 - 0}=\frac{6998.4 - 6000}{2}=\frac{998.4}{2}=499.2$.

Step2: Calculate the average rate of change between years 16 and 18

Find $f(16)$: $f(16)=6000\times1.08^{16}$.

Using a calculator, $1.08^{16}\approx3.425942643$, so $f(16)=6000\times3.425942643\approx20555.656$.

Find $f(18)$: $f(18)=6000\times1.08^{18}$.

Using a calculator, $1.08^{18}\approx4.065557303$, so $f(18)=6000\times4.065557303\approx24393.344$.

The average rate of change is $\frac{f(18)-f(16)}{18 - 16}=\frac{24393.344-20555.656}{2}=\frac{3837.688}{2}=1918.844$.

Answer:

Between years 0 and 2, the account increased by $499.2$ dollars per year. Between years 16 and 18, the account increased by $1918.844$ dollars per year. The account increased by a greater amount in the two years before Edna turned 18.