To determine when the pollution is at its highest level, we need to find the maximum value of the given function:

$P(t) = \frac{133}{1 + 0.21(t - 7)^2}$

where $t$ is the time in hours from 7 AM, with $t = 0$ corresponding to 7 AM.

Steps:

1. Find the critical points: To find the maximum, take the derivative of $P(t)$ with respect to $t$, set it to zero, and solve for $t$.

   The derivative is: $P'(t) = \frac{d}{dt} \left( \frac{133}{1 + 0.21(t - 7)^2} \right)$

2. Simplify the derivative: Use the quotient rule: $P'(t) = \frac{(1 + 0.21(t - 7)^2)(0) - 133 \cdot \frac{d}{dt}[1 + 0.21(t - 7)^2]}{(1 + 0.21(t - 7)^2)^2}$

   This simplifies to: $P'(t) = \frac{-133 \cdot (0.42(t - 7))}{(1 + 0.21(t - 7)^2)^2}$

   Factorize further: $P'(t) = \frac{-133 \cdot 0.42(t - 7)}{(1 + 0.21(t - 7)^2)^2}$

3. Set $P'(t) = 0$: $\frac{-133 \cdot 0.42(t - 7)}{(1 + 0.21(t - 7)^2)^2} = 0$

   This equation equals zero when: $t - 7 = 0 \quad \text{or} \quad t = 7$

4. Verify the maximum: At $t = 7$, the function reaches its highest value because the denominator $1 + 0.21(t - 7)^2$ is minimized.

5. Convert $t$ back to time: Since $t = 0$ corresponds to 7 AM, $t = 7$ corresponds to 2 PM.

Final Answer:

The pollution is at its highest level at 2 PM.

Would you like further clarification or detailed calculations?

Here are 5 questions for deeper exploration:

1. How is the derivative of a rational function calculated?
2. Why does the minimum denominator result in the maximum function value?
3. How would the results differ if the function were altered?
4. What is the physical significance of the derivative in this context?
5. How does this pollution model relate to real-world air quality predictions?

Tip: When analyzing rational functions, always look for where the denominator reaches its minimum (or the numerator its maximum