To determine the exponential function for Mr. Bert's investment in the form \( f(x) = a \cdot b^x \), we start with the following information:
- Initial deposit (principal) \( P = 5000 \)
- Annual interest rate \( r = 0.0375 \) (3.75%)
- Time in years \( t \)
a.) The total amount in the account after \( t \) years can be represented using the formula for compound interest:
\[ A = P(1 + r)^t \]
In this case, we must express this in the form \( f(x) = a \cdot b^x \). Here, we can take \( a = 5000 \) and can express \( b \) as:
\[ b = 1 + r = 1 + 0.0375 = 1.0375 \]
Thus, the function representing the total amount of Mr. Bert's investment account over time \( x \) (in years) is:
\[ f(x) = 5000 \cdot (1.0375)^x \]
b.) To find the total value of the investment when Mr. Bert retires in 15 years (which is after 20 + 15 = 35 years from the initial deposit), we will substitute \( x = 35 \) into our function:
\[ f(35) = 5000 \cdot (1.0375)^{35} \]
Calculating \( (1.0375)^{35} \):
\[ (1.0375)^{35} \approx 3.471 \]
Now, we can plug this back into the function:
\[ f(35) \approx 5000 \cdot 3.471 \approx 17355 \]
So, the total value of Mr. Bert's investment when he retires is approximately $17,355.
Next, to find the percent increase from the initial amount ($5,000) to the final amount ($17,355):
\[ \text{Percent Increase} = \frac{\text{Final amount} - \text{Initial amount}}{\text{Initial amount}} \times 100 \]
Substituting the values we found:
\[ \text{Percent Increase} = \frac{17355 - 5000}{5000} \times 100 \approx \frac{12355}{5000} \times 100 \approx 247.1% \]
Thus, in summary:
- The exponential function is \( f(x) = 5000 \cdot (1.0375)^x \)
- The total value of Mr. Bert's investment when he retires is approximately $17,355.
- The percent increase compared to the initial investment is approximately 247.10%.
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