Sure, let's evaluate the function [tex]\( f(x) = 450 \times (0.65)^x \)[/tex] at the specified points.
### A) Evaluate [tex]\( f(-10) \)[/tex]
To find [tex]\( f(-10) \)[/tex]:
[tex]\[ f(-10) = 450 \times (0.65)^{-10} \][/tex]
Using this exponent value, we find:
[tex]\[ f(-10) \approx 33425.58 \][/tex]
### B) Evaluate [tex]\( f(-5) \)[/tex]
To find [tex]\( f(-5) \)[/tex]:
[tex]\[ f(-5) = 450 \times (0.65)^{-5} \][/tex]
Using this exponent value, we find:
[tex]\[ f(-5) \approx 3878.34 \][/tex]
### C) Evaluate [tex]\( f(0) \)[/tex]
To find [tex]\( f(0) \)[/tex]:
[tex]\[ f(0) = 450 \times (0.65)^0 \][/tex]
Since any number to the power of zero is 1,
[tex]\[ f(0) = 450 \times 1 = 450 \][/tex]
### D) Evaluate [tex]\( f(5) \)[/tex]
To find [tex]\( f(5) \)[/tex]:
[tex]\[ f(5) = 450 \times (0.65)^5 \][/tex]
Using this exponent value, we find:
[tex]\[ f(5) \approx 52.21 \][/tex]
### E) Evaluate [tex]\( f(10) \)[/tex]
To find [tex]\( f(10) \)[/tex]:
[tex]\[ f(10) = 450 \times (0.65)^{10} \][/tex]
Using this exponent value, we find:
[tex]\[ f(10) \approx 6.06 \][/tex]
Here is a summary of the evaluations:
[tex]\[
\begin{array}{|l|c|}
\hline
A) \ f(-10) & f(-10) \approx 33425.58 \\
\hline
B) \ f(-5) & f(-5) \approx 3878.34 \\
\hline
C) \ f(0) & f(0) = 450 \\
\hline
D) \ f(5) & f(5) \approx 52.21 \\
\hline
E) \ f(10) & f(10) \approx 6.06 \\
\hline
\end{array}
\][/tex]
