The temperature of a person during a certain illness is given by the following​ equation, where T is the temperature ​(degrees​F) at time​ t, in days. Find the relative extreme points and sketch a graph of the function.
​T(t)equalsnegative 0.3 t squared plus 2.4 t plus 98.5​, 0less than or equalstless than or equals8

To find the relative extreme points of the temperature function \( T(t) = -0.3t^2 + 2.4t + 98.5 \) over the interval \( [0, 8] \), we first need to determine the critical points by taking the derivative and setting it to zero.

1. **Calculate the derivative:**
   \[
   T'(t) = -0.6t + 2.4
   \]

2. **Set the derivative equal to zero to find critical points:**
   \[
   -0.6t + 2.4 = 0
   \]
   \[
   0.6t = 2.4
   \]
   \[
   t = 4
   \]

3. **Evaluate \( T(t) \) at the critical point and endpoints:**
   - At \( t = 0 \):
     \[
     T(0) = -0.3(0)^2 + 2.4(0) + 98.5 = 98.5
     \]
   - At \( t = 4 \):
     \[
     T(4) = -0.3(4)^2 + 2.4(4) + 98.5 = -0.3(16) + 9.6 + 98.5 = -4.8 + 9.6 + 98.5 = 103.3
     \]
   - At \( t = 8 \):
     \[
     T(8) = -0.3(8)^2 + 2.4(8) + 98.5 = -0.3(64) + 19.2 + 98.5 = -19.2 + 19.2 + 98.5 = 98.5
     \]

4. **Summarize the values:**
   - \( T(0) = 98.5 \)
   - \( T(4) = 103.3 \) (relative maximum)
   - \( T(8) = 98.5 \)

The relative extreme point is at \( t = 4 \) with a maximum temperature of \( 103.3 \) °F. The function decreases before and after this point within the interval.

### Graph Sketch
To sketch the graph, plot the points:
- (0, 98.5)
- (4, 103.3)
- (8, 98.5)

The parabola opens downwards, with a vertex at \( (4, 103.3) \).

### Summary
- Relative maximum at \( t = 4 \) with \( T(4) = 103.3 \) °F.
- The temperature decreases on either side of this point within the interval \( [0, 8] \).

Would you like more details or have any questions? Here are some related questions you might consider:
1. What is the significance of finding relative extreme points in this context?
2. How would the graph change if the coefficient of \( t^2 \) were positive?
3. What does the negative coefficient of \( t^2 \) indicate about the temperature trend?
4. Can you explain the role of derivatives in finding extreme points?
5. How would you interpret the results if \( T(t) \) represented a different scenario, like a medication effect over time?

**Tip:** Always check the endpoints of your interval when looking for relative extrema!

The temperature of a person during a certain illness is given by the following equation, where T is the temperature (degrees F) at time t, in days. Find the relative extreme points and sketch a graph of the function. T(t) equals negative 0.3 t squared plus 2.4 t plus 98.5, 0 less than or equal to t less than or equal to 8.

Explore how to find relative extreme points for the quadratic temperature function T(t) = -0.3t^2 + 2.4t + 98.5 over the interval [0, 8]. The analysis reveals a maximum temperature of 103.3°F at t = 4. Ideal for high school students studying calculus.

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