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 0.2 t squared plus 2.4 t plus 98.6−0.2t2+2.4t+98.6​, 0less than or equals≤tless than or equals≤1212
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Part 1
What are the relative extreme​ points? Select the correct choice below and fill in the answer box to complete your choice

To find the relative extreme points of the function \( T(t) = -0.2t^2 + 2.4t + 98.6 \), we need to follow these steps:

1. **Find the derivative** of the function \( T(t) \).
2. **Set the derivative to zero** to find the critical points.
3. **Determine whether each critical point is a maximum or minimum** by using the second derivative test or by evaluating the sign changes of the first derivative.

### Step 1: Find the derivative of \( T(t) \)
The first derivative \( T'(t) \) is:

\[
T'(t) = \frac{d}{dt} \left( -0.2t^2 + 2.4t + 98.6 \right) = -0.4t + 2.4
\]

### Step 2: Set the derivative equal to zero
Set \( T'(t) = 0 \):

\[
-0.4t + 2.4 = 0
\]

Solving for \( t \):

\[
-0.4t = -2.4 \implies t = 6
\]

### Step 3: Determine the nature of the critical point

To classify whether \( t = 6 \) is a maximum or minimum, we can use the second derivative test. The second derivative \( T''(t) \) is:

\[
T''(t) = \frac{d}{dt} (-0.4t + 2.4) = -0.4
\]

Since \( T''(t) = -0.4 < 0 \), this indicates that \( T(t) \) has a **relative maximum** at \( t = 6 \).

### Relative Extreme Points

The relative maximum occurs at \( t = 6 \). To find the corresponding temperature \( T(6) \):

\[
T(6) = -0.2(6)^2 + 2.4(6) + 98.6
\]
\[
T(6) = -0.2(36) + 14.4 + 98.6
\]
\[
T(6) = -7.2 + 14.4 + 98.6 = 105.8
\]

So, the relative maximum point is \( (6, 105.8) \).

### Graph of the Function
To sketch the graph of \( T(t) \) over the interval \( 0 \leq t \leq 12 \), note the following key points:
- The relative maximum occurs at \( t = 6 \) with \( T(6) = 105.8 \).
- The endpoints \( T(0) = 98.6 \) and \( T(12) = 98.6 \) can be calculated directly.

Here’s a quick sketch of the graph showing the relative maximum at \( (6, 105.8) \):

```
   T(t)
    |
106 |                *
    |              / \
    |             /   \
    |            /     \
    |           /       \
    |          /         \
 99 |_________/           \________
    |_________________________________ t
    0         6                        12
```

This sketch shows the temperature peaking at 105.8°F on the sixth day and being lower at both the start and end of the period.

Do you want more details or have any questions?

### Related Questions
1. How do you find inflection points of a function?
2. How does the sign of the second derivative affect the shape of the graph?
3. What are the implications of a negative second derivative for a real-world temperature model?
4. Can there be more than one relative maximum in a quadratic function?
5. How would changing the coefficients of the function affect the location of the maximum?

### Tip
To better understand the behavior of the function, always analyze both the first and second derivatives for critical points and concavity.

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) = -0.2t^2 + 2.4t + 98.6, 0 ≤ t ≤ 12

Learn how to find the relative extreme points of the quadratic function T(t) = -0.2t^2 + 2.4t + 98.6, where T is the temperature in degrees°F at time t in days. This problem involves taking the derivative and using the second derivative test to find the relative maximum, suitable for Grades 10-12.

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