To complete the sentences, we first need to analyze the function \( f(x) = \frac{1}{9} \left( \frac{2}{3} \right)^x \) and observe how it changes as \( x \) increases by 1 unit.
Calculating the values of \( f(x) \) for the given \( x \) values:
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For \( x = -2 \): \[ f(-2) = \frac{1}{9} \left( \frac{2}{3} \right)^{-2} = \frac{1}{9} \cdot \frac{9}{4} = \frac{1}{4} \]
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For \( x = -1 \): \[ f(-1) = \frac{1}{9} \left( \frac{2}{3} \right)^{-1} = \frac{1}{9} \cdot \frac{3}{2} = \frac{1}{6} \]
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For \( x = 0 \): \[ f(0) = \frac{1}{9} \cdot 1 = \frac{1}{9} \]
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For \( x = 1 \): \[ f(1) = \frac{1}{9} \cdot \frac{2}{3} = \frac{2}{27} \]
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For \( x = 2 \): \[ f(2) = \frac{1}{9} \cdot \left( \frac{2}{3} \right)^2 = \frac{1}{9} \cdot \frac{4}{9} = \frac{4}{81} \]
Now, let's look at how \( f(x) \) changes as \( x \) increases by 1:
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From \( x = -2 \) to \( x = -1 \): \[ f(-1) - f(-2) = \frac{1}{6} - \frac{1}{4} = \frac{2}{12} - \frac{3}{12} = -\frac{1}{12} \]
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From \( x = -1 \) to \( x = 0 \): \[ f(0) - f(-1) = \frac{1}{9} - \frac{1}{6} = \frac{2}{18} - \frac{3}{18} = -\frac{1}{18} \]
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From \( x = 0 \) to \( x = 1 \): \[ f(1) - f(0) = \frac{2}{27} - \frac{1}{9} = \frac{2}{27} - \frac{3}{27} = -\frac{1}{27} \]
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From \( x = 1 \) to \( x = 2 \): \[ f(2) - f(1) = \frac{4}{81} - \frac{2}{27} = \frac{4}{81} - \frac{6}{81} = -\frac{2}{81} \]
As we can see from the calculations above, the amount by which \( f(x) \) changes decreases as \( x \) increases in each step. However, we also notice that the changes are proportional to \( f(x) \) at each step, which is characteristic of exponential decay.
Now we can fill in the blanks:
When the x-value changes by 1 unit, the function's value changes by a decreasing amount. So, \( f(x) \) demonstrates that exponential functions change by equal percentages over equal intervals.
This fits the context of exponential functions where the outputs decrease at a consistent rate relative to their current value.