Answer :
To find the relative extreme points of the temperature function [tex]\( T(t) = -0.3t^2 + 2.4t + 98.2 \)[/tex], we can follow these steps:
1. Find the Derivative:
The derivative [tex]\( T'(t) \)[/tex] represents the rate of change of the temperature with respect to time. To find it, we differentiate the function:
[tex]\[
T'(t) = \frac{d}{dt}(-0.3t^2 + 2.4t + 98.2) = -0.6t + 2.4
\][/tex]
2. Find Critical Points:
Critical points occur where the derivative is equal to zero or undefined. Since our derivative is a linear function, we find where it equals zero:
[tex]\[
-0.6t + 2.4 = 0
\][/tex]
Solving for [tex]\( t \)[/tex], we have:
[tex]\[
t = \frac{2.4}{0.6} = 4
\][/tex]
3. Evaluate the Temperature at the Critical Point:
Substitute [tex]\( t = 4 \)[/tex] back into the original temperature function to get the temperature at this critical point:
[tex]\[
T(4) = -0.3(4^2) + 2.4(4) + 98.2 = -0.3(16) + 9.6 + 98.2 = -4.8 + 9.6 + 98.2 = 103
\][/tex]
4. Analyze the Second Derivative:
The second derivative tells us about the concavity of the function. To check the nature of the critical point, find the second derivative:
[tex]\[
T''(t) = \frac{d^2}{dt^2}(-0.3t^2 + 2.4t + 98.2) = -0.6
\][/tex]
Since [tex]\( T''(t) = -0.6 \)[/tex] and it is negative, the function is concave down at [tex]\( t = 4 \)[/tex]. Therefore, the critical point at [tex]\( t = 4 \)[/tex] is a maximum.
5. Conclusion:
The relative maximum temperature occurs at [tex]\( t = 4 \)[/tex] days, with a temperature of 103°F.
Finally, when sketching the graph, note that the temperature starts at [tex]\( 98.2 \)[/tex] at [tex]\( t = 0 \)[/tex], rises to a maximum of [tex]\( 103 \)[/tex] at [tex]\( t = 4 \)[/tex], and then decreases. Make sure to plot these key points to help illustrate the behavior of the temperature function over the interval [tex]\( 0 \leq t \leq 8 \)[/tex].
1. Find the Derivative:
The derivative [tex]\( T'(t) \)[/tex] represents the rate of change of the temperature with respect to time. To find it, we differentiate the function:
[tex]\[
T'(t) = \frac{d}{dt}(-0.3t^2 + 2.4t + 98.2) = -0.6t + 2.4
\][/tex]
2. Find Critical Points:
Critical points occur where the derivative is equal to zero or undefined. Since our derivative is a linear function, we find where it equals zero:
[tex]\[
-0.6t + 2.4 = 0
\][/tex]
Solving for [tex]\( t \)[/tex], we have:
[tex]\[
t = \frac{2.4}{0.6} = 4
\][/tex]
3. Evaluate the Temperature at the Critical Point:
Substitute [tex]\( t = 4 \)[/tex] back into the original temperature function to get the temperature at this critical point:
[tex]\[
T(4) = -0.3(4^2) + 2.4(4) + 98.2 = -0.3(16) + 9.6 + 98.2 = -4.8 + 9.6 + 98.2 = 103
\][/tex]
4. Analyze the Second Derivative:
The second derivative tells us about the concavity of the function. To check the nature of the critical point, find the second derivative:
[tex]\[
T''(t) = \frac{d^2}{dt^2}(-0.3t^2 + 2.4t + 98.2) = -0.6
\][/tex]
Since [tex]\( T''(t) = -0.6 \)[/tex] and it is negative, the function is concave down at [tex]\( t = 4 \)[/tex]. Therefore, the critical point at [tex]\( t = 4 \)[/tex] is a maximum.
5. Conclusion:
The relative maximum temperature occurs at [tex]\( t = 4 \)[/tex] days, with a temperature of 103°F.
Finally, when sketching the graph, note that the temperature starts at [tex]\( 98.2 \)[/tex] at [tex]\( t = 0 \)[/tex], rises to a maximum of [tex]\( 103 \)[/tex] at [tex]\( t = 4 \)[/tex], and then decreases. Make sure to plot these key points to help illustrate the behavior of the temperature function over the interval [tex]\( 0 \leq t \leq 8 \)[/tex].
