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The function provided is \( Q = f(t) = 100t e^{-0.5t} \). We need to find \( f(6) \), \( f'(6) \), and interpret the results.

### 1. Compute \( f(6) \)
To find \( f(6) \), substitute \( t = 6 \) into the function:

\[ f(6) = 100 \cdot 6 \cdot e^{-0.5 \cdot 6} \]

### 2. Compute \( f'(t) \) and \( f'(6) \)
To find the rate of change of the drug quantity at time \( t \), we need the derivative \( f'(t) \). We'll use the product rule, where if \( f(t) = uv \), then \( f'(t) = u'v + uv' \). Here, \( u = 100t \) and \( v = e^{-0.5t} \). The derivatives are:

- \( u' = 100 \)
- \( v' = \frac{d}{dt} [e^{-0.5t}] = -0.5 e^{-0.5t} \)

So,
\[ f'(t) = 100 \cdot e^{-0.5t} + 100t \cdot (-0.5 e^{-0.5t}) = 100e^{-0.5t}(1 - 0.5t) \]

Substitute \( t = 6 \) into \( f'(t) \) to find \( f'(6) \):

\[ f'(6) = 100e^{-0.5 \cdot 6}(1 - 0.5 \cdot 6) \]

Now, let's calculate these values.The results are:
- \( f(6) = 29.87 \) mg
- \( f'(6) = -9.96 \) mg per hour

### Interpretation
Six hours after the drug was administered, the quantity of the drug in the body is approximately 29.87 mg, and the quantity is decreasing at a rate of about 9.96 mg per hour.

This indicates that 6 hours post-administration, the drug's concentration in the body is diminishing, reflecting its absorption or metabolism process.

Would you like more details or have any questions about this? Here are some related questions you might consider:

1. How would the quantity change if the decay rate in the exponent was different?
2. What is the maximum quantity of the drug in the body and at what time does it occur?
3. How does the rate of change of the drug's quantity vary over time?
4. What are the implications of this rate of change for dosing schedules?
5. Can you predict when the drug's quantity will fall below a specific threshold?

**Tip:** For a more comprehensive understanding, graphing the function \( f(t) \) and its derivative \( f'(t) \) might visually demonstrate how the drug's concentration changes over time.

The quantity of a drug, Q mg, present in the body t hours after an injection of the drug is given is Q = f(t) = 100te^{-0.5t}. Find f(6), f'(6), and interpret the result. Round your answers to two decimal places.

Learn how to solve an exponential decay problem related to drug concentration in the body using calculus. We find f(6) = 29.87 mg and f'(6) = -9.96 mg per hour, showing how the quantity decreases over time.

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