Write an exponential decay function to model the situation. Compare the average rates of change over the given intervals. initial value: 49 decay factor: 0.8 $$ 1 \leq x \leq 3 $$ and $$ 4 \leq x \leq 8 $$ The exponential decay function to model the situation is $$ f(x)=49=0.8^{x} $$. The average rate of change over $$ 1 \leq x \leq 3 $$ is $$ \square $$. The average rate of change over $$ 4 \leq x \leq 8 $$ is $$ \square $$. The rate of change $$ \begin{array}{l} ext { (Round to three decimal places as needed.) }\end{array} $$ $$ \begin{array}{l} ext { as increases. }\end{array} $$

Question

Write an exponential decay function to model the situation. Compare the average rates of change over the given intervals. initial value: 49 decay factor: 0.8 $$ 1 \leq x \leq 3 $$ and $$ 4 \leq x \leq 8 $$ The exponential decay function to model the situation is $$ f(x)=49=0.8^{x} $$. The average rate of change over $$ 1 \leq x \leq 3 $$ is $$ \square $$. The average rate of change over $$ 4 \leq x \leq 8 $$ is $$ \square $$. The rate of change $$ \begin{array}{l}	ext { (Round to three decimal places as needed.) }\end{array} $$ $$ \begin{array}{l}	ext { as increases. }\end{array} $$

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To model the given situation using an exponential decay function and compare the average rates of change over the specified intervals, follow these steps:

### 1. **Exponential Decay Function**

The general form of an exponential decay function is:

$$ f(x) = A \times b^x $$

Where:
- $$ A $$ is the initial value.
- $$ b $$ is the decay factor (with $$ 0 < b < 1 $$).

Given:
- **Initial value ($$ A $$)**: 49
- **Decay factor ($$ b $$)**: 0.8

**Therefore, the exponential decay function is:**

$$ f(x) = 49 \times 0.8^x $$

### 2. **Average Rate of Change**

The **average rate of change** of a function $$ f(x) $$ over the interval $$ [a, b] $$ is calculated using the formula:

$$ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} $$

Let's compute this for both intervals.

---

#### a. **Interval $$ 1 \leq x \leq 3 $$**

1. **Calculate $$ f(1) $$ and $$ f(3) $$:**
   $$
   \begin{align*}
   f(1) &= 49 \times 0.8^1 = 49 \times 0.8 = 39.2 \
   f(3) &= 49 \times 0.8^3 = 49 \times 0.512 = 25.088 \
   \end{align*}
   $$

2. **Compute the Average Rate of Change:**
   $$
   \frac{f(3) - f(1)}{3 - 1} = \frac{25.088 - 39.2}{2} = \frac{-14.112}{2} = -7.056
   $$

**Average Rate of Change over $$ 1 \leq x \leq 3 $$ is $$ \boxed{-7.056} $$.**

---

#### b. **Interval $$ 4 \leq x \leq 8 $$**

1. **Calculate $$ f(4) $$ and $$ f(8) $$:**
   $$
   \begin{align*}
   f(4) &= 49 \times 0.8^4 = 49 \times 0.4096 = 20.0816 \
   f(8) &= 49 \times 0.8^8 = 49 \times 0.16777216 \approx 8.221 \
   \end{align*}
   $$

2. **Compute the Average Rate of Change:**
   $$
   \frac{f(8) - f(4)}{8 - 4} = \frac{8.221 - 20.0816}{4} = \frac{-11.8606}{4} \approx -2.965
   $$

**Average Rate of Change over $$ 4 \leq x \leq 8 $$ is $$ \boxed{-2.965} $$.**

---

### 3. **Comparison of Rates of Change**

- **From $$ 1 \leq x \leq 3 $$:** The average rate of change is **-7.056**.
- **From $$ 4 \leq x \leq 8 $$:** The average rate of change is **-2.965**.

**Interpretation:**
- The average rate of change becomes **less negative** as $$ x $$ increases. This means that while the quantity is still decreasing, it is doing so at a **slower rate** in the interval $$ 4 \leq x \leq 8 $$ compared to $$ 1 \leq x \leq 3 $$.

### 4. **Final Summary**

- **Exponential Decay Function:** $$ f(x) = 49 \times 0.8^x $$
- **Average Rate of Change:**
  - $$ 1 \leq x \leq 3 $$: **-7.056**
  - $$ 4 \leq x \leq 8 $$: **-2.965**
- **Conclusion:** The rate of change **increases** (becomes less negative) as $$ x $$ increases.
The exponential decay function is $$ f(x) = 49 \times 0.8^x $$. - **Average Rate of Change over $$ 1 \leq x \leq 3 $$:** -7.056 - **Average Rate of Change over $$ 4 \leq x \leq 8 $$:** -2.965 As $$ x $$ increases, the rate of change **increases** (becomes less negative).

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Solución

1. Exponential Decay Function

2. Average Rate of Change

a. Interval

b. Interval

3. Comparison of Rates of Change

4. Final Summary

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