Answer:
Option d. [tex]f(x)=\$1,000(1+0.0355)^{x}[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=x\ years\\ P=\$1,000\\ r=0.0355\\n=1[/tex]
substitute in the formula above
[tex]A=\$1,000(1+\frac{0.0355}{1})^{1*x}[/tex]
[tex]A=\$1,000(1+0.0355)^{x}[/tex]
Convert to function notation
[tex]f(x)=\$1,000(1+0.0355)^{x}[/tex]
