The number of visitors P to a website in a given week over a​ 1-year period is given by Upper P left parenthesis t right parenthesis equals 115 plus left parenthesis t minus 85 right parenthesis e Superscript 0.02 tP(t)=115+(t−85)e0.02t​, where t is the week and 1less than or equals≤tless than or equals≤52.
​a) Over what interval of time during the​ 1-year period is the number of visitors​ decreasing?
​b) Over what interval of time during the​ 1-year period is the number of visitors​ increasing?
​c) Find the critical​ point, and interpret its meaning.
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Part 1
​a) The number of visitors is decreasing over the interval left parenthesis 1 comma 35 right parenthesis(1,35).
​(Simplify your answer. Type integers or decimals rounded to three decimal places as needed. Type your answer in interval​ notation.)
Part 2
​b) The number of visitors is increasing over the interval left parenthesis 35 comma 52 right parenthesis(35,52).
​(Simplify your answer. Type integers or decimals rounded to three decimal places as needed. Type your answer in interval​ notation.)
Part 3
​c) The critical point is left parenthesis 35 comma 14.312 right parenthesis(35,14.312). The critical point means that the number of visitors was
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in week enter your response here at approximately enter your response here visitors.

Let's break down the problem and solve it step by step, using calculus.

### Given:
The number of visitors to a website as a function of time \( t \) (in weeks) is given by:

\[
P(t) = 115 + (t - 85)e^{0.02t}
\]
where \( t \) is between 1 and 52 weeks.

### Part 1: Find the interval where the number of visitors is decreasing.

To find where the number of visitors is decreasing, we need to look at the derivative \( P'(t) \). Decreasing occurs when \( P'(t) < 0 \).

1. **Find \( P'(t) \)** (the first derivative of \( P(t) \)):

\[
P(t) = 115 + (t - 85)e^{0.02t}
\]
Using the product rule:
\[
P'(t) = \frac{d}{dt}\left[(t - 85)e^{0.02t}\right]
\]
By applying the product rule, we get:
\[
P'(t) = (1)e^{0.02t} + (t - 85)\left(0.02e^{0.02t}\right)
\]
Simplifying:
\[
P'(t) = e^{0.02t} + 0.02(t - 85)e^{0.02t}
\]
Factor out \( e^{0.02t} \):
\[
P'(t) = e^{0.02t}(1 + 0.02(t - 85))
\]
Simplify the expression inside the parentheses:
\[
P'(t) = e^{0.02t}(0.02t - 0.7)
\]

2. **Set \( P'(t) = 0 \) to find critical points**:

\[
e^{0.02t}(0.02t - 0.7) = 0
\]
Since \( e^{0.02t} \neq 0 \), solve \( 0.02t - 0.7 = 0 \):
\[
0.02t = 0.7
\]
\[
t = 35
\]

So, the critical point occurs at \( t = 35 \).

3. **Determine intervals of increase and decrease**:

- For \( t < 35 \), let's check the sign of \( P'(t) \). Pick \( t = 1 \):
  \[
  P'(1) = e^{0.02(1)}(0.02(1) - 0.7) = e^{0.02}(-0.68) < 0
  \]
  So, \( P'(t) < 0 \) for \( 1 \leq t < 35 \), meaning the number of visitors is decreasing.

- For \( t > 35 \), pick \( t = 40 \):
  \[
  P'(40) = e^{0.02(40)}(0.02(40) - 0.7) = e^{0.8}(0.8 - 0.7) > 0
  \]
  So, \( P'(t) > 0 \) for \( t > 35 \), meaning the number of visitors is increasing.

Thus, the number of visitors is **decreasing** on the interval \( (1, 35) \) and **increasing** on \( (35, 52) \).

### Part 2: Find the interval where the number of visitors is increasing.

As determined earlier, the number of visitors is increasing on the interval \( (35, 52) \).

### Part 3: Find the critical point and interpret its meaning.

The critical point is \( t = 35 \). To find the number of visitors at this critical point, substitute \( t = 35 \) into the original function \( P(t) \):

\[
P(35) = 115 + (35 - 85)e^{0.02(35)}
\]
\[
P(35) = 115 + (-50)e^{0.7}
\]
Using \( e^{0.7} \approx 2.01375 \):
\[
P(35) = 115 - 50(2.01375) = 115 - 100.6875 = 14.3125
\]

So, the critical point is \( (35, 14.312) \).

### Interpretation:
The critical point \( (35, 14.312) \) means that in week 35, the number of visitors was approximately 14.312. This is the lowest point in the number of visitors during the year, after which the number of visitors starts to increase again.

### Summary:
a) The number of visitors is decreasing on the interval \( (1, 35) \).

b) The number of visitors is increasing on the interval \( (35, 52) \).

c) The critical point is \( (35, 14.312) \), meaning that in week 35, the number of visitors was at its lowest, approximately 14.312 visitors.

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Do you want more details on any part of the explanation? 

Here are five related questions to expand on this topic:
1. How do you determine whether a function is increasing or decreasing from its derivative?
2. What is the significance of critical points in real-world applications like this one?
3. How can we use second derivatives to confirm whether a critical point is a minimum or maximum?
4. Why do exponential functions play such a crucial role in growth and decay problems?
5. What happens to the visitor function as \( t \) approaches the endpoints of the domain?

**Tip:** When analyzing functions for increase or decrease, always start by finding the first derivative and then determine where it changes sign!

The number of visitors P to a website in a given week over a 1-year period is given by P(t) = 115 + (t − 85)e^0.02t, where t is the week and 1 ≤ t ≤ 52. a) Over what interval of time during the 1-year period is the number of visitors decreasing? b) Over what interval of time during the 1-year period is the number of visitors increasing? c) Find the critical point, and interpret its meaning.

Explore the number of website visitors as a function of time using calculus. Learn to identify intervals of increase and decrease by solving for the first derivative, and find critical points in an exponential function. Suitable for Grades 11-12.

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