To solve this problem, we need to analyze the function $P(t) = 115 + (t - 85)e^{0.02t}$, which represents the number of visitors $P$ to a website over time, where $t$ represents the week and $1 \leq t \leq 52$.

Part (a) & (b): Determining the intervals of increase and decrease

To find when the number of visitors is increasing or decreasing, we need to determine the derivative of the function $P(t)$, $P'(t)$, and analyze its sign (positive or negative).

1. Find $P'(t)$: To differentiate $P(t) = 115 + (t - 85)e^{0.02t}$, apply the product rule to the term $(t - 85)e^{0.02t}$.

   $P'(t) = \frac{d}{dt}\left[115 + (t - 85)e^{0.02t}\right]$ Since the derivative of 115 is 0, we focus on the derivative of $(t - 85)e^{0.02t}$.

   Using the product rule: $P'(t) = (t - 85) \cdot \frac{d}{dt}\left[e^{0.02t}\right] + e^{0.02t} \cdot \frac{d}{dt}(t - 85)$ The derivative of $e^{0.02t}$ is $0.02e^{0.02t}$, and the derivative of $t - 85$ is 1, so:

   $P'(t) = (t - 85) \cdot 0.02e^{0.02t} + e^{0.02t}$ Simplifying: $P'(t) = e^{0.02t} \left[ (t - 85) \cdot 0.02 + 1 \right]$ $P'(t) = e^{0.02t} \left[ 0.02t - 1.7 + 1 \right]$ $P'(t) = e^{0.02t} \left[ 0.02t - 0.7 \right]$

2. Find the critical point (where $P'(t) = 0$):

   For the critical points, set $P'(t) = 0$:

   $e^{0.02t} (0.02t - 0.7) = 0$

   Since $e^{0.02t} \neq 0$, we solve:

   $0.02t - 0.7 = 0$ $0.02t = 0.7$ $t = \frac{0.7}{0.02} = 35$

   The critical point occurs at $t = 35$.

3. Determine the sign of $P'(t)$:

   To determine if the function is increasing or decreasing, we need to test values of $t$ around the critical point $t = 35$:

   • For $t < 35$, let's pick $t = 30$: $P'(30) = e^{0.02(30)} (0.02(30) - 0.7) = e^{0.6} \cdot (0.6 - 0.7) < 0$ So, $P'(t) < 0$, meaning $P(t)$ is decreasing for $t < 35$.

   • For $t > 35$, let's pick $t = 40$: $P'(40) = e^{0.02(40)} (0.02(40) - 0.7) = e^{0.8} \cdot (0.8 - 0.7) > 0$ So, $P'(t) > 0$, meaning $P(t)$ is increasing for $t > 35$.

Summary of intervals:

• The number of visitors is decreasing for $1 \leq t < 35$.
• The number of visitors is increasing for $35 < t \leq 52$.

Part (c): Critical point interpretation

The critical point occurs at $t = 35$. This means that during the 35th week, the number of visitors reaches a minimum. After this week, the number of visitors begins to increase.

Conclusion:

• a) The number of visitors is decreasing during $1 \leq t < 35$.
• b) The number of visitors is increasing during $35 < t \leq 52$.
• c) The critical point is $t = 35$, which represents the week when the number of visitors is at its lowest point.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How can we verify if $t = 35$ is a minimum point using the second derivative test?
2. How does the value of $e^{0.02t}$ influence the behavior of the function?
3. What happens to the visitor count at $t = 52$?
4. Can we find an explicit expression for the total number of visitors over the entire year?
5. How would the intervals of increase and decrease change if the exponent in the original equation were different?

Tip: Critical points occur where the derivative is zero or undefined, and they often represent local maxima, minima, or points of inflection. Always test the derivative around these points!