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The given sales function for a new high-tech item is:

\[
S(t) = 106 - 80e^{-0.4t},
\]

where \( S(t) \) represents the sales (in thousands) and \( t \) is the time in years. We are asked to find the rate of change of sales, which is the derivative of \( S(t) \), and analyze it at different time points.

### Step 1: Find the derivative of \( S(t) \)
The rate of change of sales is the derivative \( S'(t) \), so let's differentiate:

\[
S'(t) = \frac{d}{dt} \left( 106 - 80e^{-0.4t} \right).
\]

The derivative of the constant \( 106 \) is zero, and we need to apply the chain rule to the second term:

\[
S'(t) = 0 - 80 \cdot \frac{d}{dt} \left( e^{-0.4t} \right).
\]

The derivative of \( e^{-0.4t} \) is \( -0.4e^{-0.4t} \), so:

\[
S'(t) = -80 \cdot (-0.4e^{-0.4t}),
\]
\[
S'(t) = 32e^{-0.4t}.
\]

This is the rate of change of sales over time.

### Step 2: Find \( S'(t) \) at specific times

#### a) After 1 year (\( t = 1 \)):
Substitute \( t = 1 \) into the derivative \( S'(t) = 32e^{-0.4t} \):

\[
S'(1) = 32e^{-0.4 \cdot 1} = 32e^{-0.4}.
\]

Using \( e^{-0.4} \approx 0.67032 \):

\[
S'(1) \approx 32 \cdot 0.67032 = 21.45.
\]

So, the rate of change of sales after 1 year is approximately **21.45 thousand units per year**.

#### b) After 5 years (\( t = 5 \)):
Substitute \( t = 5 \) into the derivative:

\[
S'(5) = 32e^{-0.4 \cdot 5} = 32e^{-2}.
\]

Using \( e^{-2} \approx 0.13534 \):

\[
S'(5) \approx 32 \cdot 0.13534 = 4.33.
\]

So, the rate of change of sales after 5 years is approximately **4.33 thousand units per year**.

### Step 3: Analyze the rate of change as time goes on

As \( t \) increases, \( e^{-0.4t} \) becomes smaller and approaches zero. Therefore, \( S'(t) = 32e^{-0.4t} \) also decreases over time, approaching zero.

Thus, as time goes on, the rate of change of sales **decreases** and approaches zero, meaning sales are growing more slowly over time.

### Step 4: Does the rate of change of sales ever equal zero?

Since the derivative is \( S'(t) = 32e^{-0.4t} \), and \( e^{-0.4t} \) never reaches zero (it only approaches zero as \( t \to \infty \)), the rate of change of sales **never exactly equals zero** at any finite time. It gets closer and closer to zero as \( t \) increases, but it never actually reaches zero.

### Summary of Answers:
- a) After 1 year, the rate of change of sales is approximately **21.45 thousand units/year**.
- b) After 5 years, the rate of change of sales is approximately **4.33 thousand units/year**.
- c) The rate of change of sales decreases over time as \( t \) increases.
- d) The rate of change of sales **never equals zero** at any finite time.

Would you like further clarification on any of these parts or more details?

---

Here are 5 related questions to consider:

1. How would the rate of change of sales change if the exponent \( -0.4t \) were different?
2. What happens to the sales \( S(t) \) itself as \( t \) goes to infinity?
3. Can you interpret this behavior of sales in a real-world context (e.g., product lifecycle)?
4. How would you find the time at which the rate of change is half of its initial value?
5. What would happen if the sales function included an additional linear term like \( 10t \)?

**Tip:** When dealing with exponential functions in derivatives, always consider how the exponential term behaves as \( t \) grows large, since it often dictates long-term behavior.

The sales of a new high-tech item (in thousands) are given by S(t) = 106 - 80e^-0.4t, where t represents time in years. Find the rate of change of sales at each time. a) After 1 year. b) After 5 years. c) What is happening to the rate of change of sales as time goes on? d) Does the rate of change of sales ever equal zero?

Explore the rate of change of sales for a high-tech item using exponential functions. This problem involves finding the derivative of the sales function, S(t) = 106 - 80e^-0.4t, and analyzing its behavior after 1 year, 5 years, and as time progresses. Learn how the rate of change approaches zero over time, with step-by-step calculations. Suitable for advanced high school or college students studying calculus.

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