Let's evaluate the function [tex]\( f(x) = 2.5 e^{-0.04 x} - 0.2 x - 1.2 \)[/tex] at the specified points: the lower bound, the average of the bounds, and the upper bound.
### Lower Bound (4)
1. First, substitute [tex]\( x = 4 \)[/tex] into the function:
[tex]\[
f(4) = 2.5 e^{-0.04 \cdot 4} - 0.2 \cdot 4 - 1.2
\][/tex]
2. Calculate the value of the exponential part [tex]\( e^{-0.16} \)[/tex] and then evaluate the rest of the expression:
[tex]\[
f(4) = 2.5 e^{-0.16} - 0.8 - 1.2 \approx 0.130359472415528
\][/tex]
### Average of the Bounds (4.5)
1. Next, substitute [tex]\( x = 4.5 \)[/tex] into the function:
[tex]\[
f(4.5) = 2.5 e^{-0.04 \cdot 4.5} - 0.2 \cdot 4.5 - 1.2
\][/tex]
2. Calculate [tex]\( e^{-0.18} \)[/tex] and then evaluate the rest of the expression:
[tex]\[
f(4.5) = 2.5 e^{-0.18} - 0.9 - 1.2 \approx -0.0118244714718200
\][/tex]
### Upper Bound (5)
1. Finally, substitute [tex]\( x = 5 \)[/tex] into the function:
[tex]\[
f(5) = 2.5 e^{-0.04 \cdot 5} - 0.2 \cdot 5 - 1.2
\][/tex]
2. Calculate [tex]\( e^{-0.20} \)[/tex] and then evaluate the rest of the expression:
[tex]\[
f(5) = 2.5 e^{-0.20} - 1 - 1.2 \approx -0.153173117305045
\][/tex]
### Summary
Let's compile these results in the table:
[tex]\[
\begin{tabular}{|l|l|}
\hline
\multicolumn{1}{|c|}{x} & 2.5 e^{-0.04 x} - 0.2 x - 1.2 \\
\hline
Lower bound (4) & 0.130359472415528 \\
\hline
Average of the bounds (4.5) & -0.0118244714718200 \\
\hline
Upper bound (5) & -0.153173117305045 \\
\hline
\end{tabular}
\][/tex]
