To solve the problem, we need to find the values of the function [tex]\( f(x) = -2 \cdot (0.5)^x \)[/tex] for the given values of [tex]\( x \)[/tex]. These values of [tex]\( x \)[/tex] are [tex]\(-2, -1, 0, 1,\)[/tex] and [tex]\(2\)[/tex]. We'll substitute each value of [tex]\( x \)[/tex] into the function and solve for [tex]\( f(x) \)[/tex].
### Step-by-Step Calculation:
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = -2 \cdot (0.5)^{-2}
\][/tex]
[tex]\[
(0.5)^{-2} = \left( \frac{1}{2} \right)^{-2} = 2^2 = 4
\][/tex]
[tex]\[
f(-2) = -2 \cdot 4 = -8.0
\][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = -2 \cdot (0.5)^{-1}
\][/tex]
[tex]\[
(0.5)^{-1} = \left( \frac{1}{2} \right)^{-1} = 2
\][/tex]
[tex]\[
f(-1) = -2 \cdot 2 = -4.0
\][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = -2 \cdot (0.5)^0
\][/tex]
[tex]\[
(0.5)^0 = 1 \quad \text{(Any number raised to the power of zero is 1)}
\][/tex]
[tex]\[
f(0) = -2 \cdot 1 = -2.0
\][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = -2 \cdot (0.5)^1
\][/tex]
[tex]\[
(0.5)^1 = 0.5
\][/tex]
[tex]\[
f(1) = -2 \cdot 0.5 = -1.0
\][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = -2 \cdot (0.5)^2
\][/tex]
[tex]\[
(0.5)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4} = 0.25
\][/tex]
[tex]\[
f(2) = -2 \cdot 0.25 = -0.5
\][/tex]
### Final Results:
Let's summarize in the table:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$f(x)$ & -8.0 & -4.0 & -2.0 & -1.0 & -0.5 \\
\hline
\end{tabular}
\][/tex]
Thus, the function values for the given [tex]\( x \)[/tex] values are:
- When [tex]\( x = -2 \)[/tex], [tex]\( f(x) = -8.0 \)[/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = -4.0 \)[/tex]
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -2.0 \)[/tex]
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -1.0 \)[/tex]
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = -0.5 \)[/tex]
