Sure, let's take this step-by-step:
1. Finding [tex]\( f(0) \)[/tex]:
Given the formula for the amount in the account:
[tex]\[ A = f(t) = 11100 e^{0.022 t} \][/tex]
To find [tex]\( f(0) \)[/tex], we substitute [tex]\( t = 0 \)[/tex]:
[tex]\[ f(0) = 11100 e^{0.022 \cdot 0} \][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[ f(0) = 11100 \times 1 = 11100 \][/tex]
So,
[tex]\[ f(0) = \$11100.00 \][/tex]
2. Finding the amount in the account after 9 years:
To find the amount after 9 years, substitute [tex]\( t = 9 \)[/tex] into the formula:
[tex]\[ A = 11100 e^{0.022 \cdot 9} \][/tex]
Using the given information:
[tex]\[ A \approx 13530.48 \][/tex]
Therefore, to the nearest cent, the amount in the account after 9 years is:
[tex]\[ \$13530.48 \][/tex]
3. Determining the time it will take for the investment to grow to \[tex]$12400:
We need to find \( t \) when the amount \( A = 12400 \):
\[ 12400 = 11100 e^{0.022 t} \]
To isolate \( t \), we first divide both sides by 11100:
\[ \frac{12400}{11100} = e^{0.022 t} \]
\[ \frac{12400}{11100} \approx 1.117117 \]
Take the natural logarithm (ln) of both sides:
\[ \ln(1.117117) = 0.022 t \]
Solve for \( t \):
\[ t = \frac{\ln(1.117117)}{0.022} \]
Using the given information:
\[ t \approx 5.0 \]
Therefore, it will take approximately:
\[ 5.0 \] years for the investment to grow to \$[/tex]12400.
