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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.

5 answers

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.

how much does the function value change by every unit?

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 \( 33.33% \).

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

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.