Question

This table shows some values of the function f(x)=1/9(2/3)^x
x
–
2
–
1 0 1 2
f(x)
1/4
1/6
1/9
2/27
4/81
Complete the sentences.
When the x-value changes by 1 unit, the function's value changes by ___. So, f(x) demonstrates that ___ functions change by equal ___ over equal intervals.

Answer 1

how much does the function value change by every unit?

Answer 2

is it "a factor of 2/3" or is it "2/3"

Answer 3

To complete the sentences, let's analyze the function \( f(x) = \frac{1}{9} \left( \frac{2}{3} \right)^x \) based on the given values.

### Given values of \( f(x) \):
- \( f(-2) = \frac{1}{4} \)
- \( f(-1) = \frac{1}{6} \)
- \( f(0) = \frac{1}{9} \)
- \( f(1) = \frac{2}{27} \)
- \( f(2) = \frac{4}{81} \)

Next, we will calculate the changes in function values when \( x \) changes by 1 unit.

1. Change from \( x = -2 \) to \( x = -1 \):
\[
\Delta f = f(-1) - f(-2) = \frac{1}{6} - \frac{1}{4} = \frac{2}{12} - \frac{3}{12} = -\frac{1}{12}
\]

2. Change from \( x = -1 \) to \( x = 0 \):
\[
\Delta f = f(0) - f(-1) = \frac{1}{9} - \frac{1}{6} = \frac{2}{18} - \frac{3}{18} = -\frac{1}{18}
\]

3. Change from \( x = 0 \) to \( x = 1 \):
\[
\Delta f = f(1) - f(0) = \frac{2}{27} - \frac{1}{9} = \frac{2}{27} - \frac{3}{27} = -\frac{1}{27}
\]

4. Change from \( x = 1 \) to \( x = 2 \):
\[
\Delta f = f(2) - f(1) = \frac{4}{81} - \frac{2}{27} = \frac{4}{81} - \frac{6}{81} = -\frac{2}{81}
\]

### Pattern:
As we observe, \( f(x) \) indeed decreases, but the changes are not equal. However, we can say that it decreases at a consistent rate relative to its previous value, which is characteristic of exponential decay functions.

### Completed Sentences:
When the x-value changes by 1 unit, the function's value changes by **a decreasing amount that varies but follows a consistent pattern**. So, \( f(x) \) demonstrates that **exponential** functions change by equal **percentages (or ratios)** over equal intervals.

Answer 4

To determine how much the function value \( f(x) = \frac{1}{9} \left( \frac{2}{3} \right)^x \) changes for every unit increase in \( x \), we can calculate the percentage change in \( f(x) \) as \( x \) increases by 1.

We can compute the ratio of the function values at successive \( x \) values, or we can look at the actual changes in function values between them.

### Calculation of function values and changes:

1. **For \( x = -2 \) to \( x = -1 \)**:
- \( f(-2) = \frac{1}{4} \)
- \( f(-1) = \frac{1}{6} \)
- Change: \( \Delta f = f(-1) - f(-2) = \frac{1}{6} - \frac{1}{4} = -\frac{1}{12} \)

2. **For \( x = -1 \) to \( x = 0 \)**:
- \( f(0) = \frac{1}{9} \)
- Change: \( \Delta f = f(0) - f(-1) = \frac{1}{9} - \frac{1}{6} = -\frac{1}{18} \)

3. **For \( x = 0 \) to \( x = 1 \)**:
- \( f(1) = \frac{2}{27} \)
- Change: \( \Delta f = f(1) - f(0) = \frac{2}{27} - \frac{1}{9} = \frac{2}{27} - \frac{3}{27} = -\frac{1}{27} \)

4. **For \( x = 1 \) to \( x = 2 \)**:
- \( f(2) = \frac{4}{81} \)
- Change: \( \Delta f = f(2) - f(1) = \frac{4}{81} - \frac{2}{27} = \frac{4}{81} - \frac{6}{81} = -\frac{2}{81} \)

### Summary of Changes in Function Values:
- \( \Delta f \) from \( x = -2 \) to \( x = -1 \): \( -\frac{1}{12} \)
- \( \Delta f \) from \( x = -1 \) to \( x = 0 \): \( -\frac{1}{18} \)
- \( \Delta f \) from \( x = 0 \) to \( x = 1 \): \( -\frac{1}{27} \)
- \( \Delta f \) from \( x = 1 \) to \( x = 2 \): \( -\frac{2}{81} \)

### Rate of Change:
The change in the function's value is not constant but rather decreases every time \( x \) increases by 1. However, the function's value does decrease proportionally relative to its previous value, characteristic of exponential decay.

#### Percentage or Ratio:
1. **To express the amount the function value changes by in a more general sense**, you can calculate the ratio of \( f(x + 1) \) to \( f(x) \):
- The ratio for any point is:
\[
\text{Ratio} = f(x+1) / f(x) = \frac{1}{9} \left( \frac{2}{3} \right)^{x+1} / \frac{1}{9} \left( \frac{2}{3} \right)^x = \frac{2}{3}
\]

So, every time \( x \) increases by 1, the function \( f(x) \) is multiplied by \( \frac{2}{3} \), indicating that it retains \( \frac{2}{3} \) (or approximately 66.67%) of its previous value, representing a decrease of about 33.33%.

### Conclusion:
As \( x \) increases by 1 unit, the function value changes by **multiplying the previous value by \( \frac{2}{3} \)**, which corresponds to a decrease of approximately \( 33.33\% \).

Answer 5

The correct expression would be **"a factor of \( \frac{2}{3} \)"**. This terminology emphasizes that the function value is multiplied by \( \frac{2}{3} \) each time \( x \) increases by 1.

### Explanation:
- When we say that a function changes by "a factor of \( \frac{2}{3} \)", it implies that if you have a value \( f(x) \), the new value \( f(x+1) \) can be calculated as:
\[
f(x+1) = \frac{2}{3} \cdot f(x)
\]

This means the function’s value shrinks to \( \frac{2}{3} \) of its previous value with each unit increase in \( x \).

### In Context:
So, you can say:
- "When \( x \) increases by 1 unit, the function’s value is reduced by **a factor of \( \frac{2}{3} \)**."
- This also means that the function’s value retains \( \frac{2}{3} \) of its previous value.

This terminology appropriately conveys the process of the function's behavior over these changes.