Answer :
To find the relative extreme points of the function [tex]\( T(t) = -0.1t^2 + 1.4t + 98.5 \)[/tex], which represents the temperature at time [tex]\( t \)[/tex] during an illness, we need to follow these steps:
1. Determine the Derivative:
The first step is to find the derivative of [tex]\( T(t) \)[/tex]. The derivative, [tex]\( \frac{dT}{dt} \)[/tex], gives the rate of change of temperature with respect to time and helps us find the critical points where the temperature might be at an extreme (either a maximum or a minimum).
The derivative of the function [tex]\( T(t) = -0.1t^2 + 1.4t + 98.5 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[
\frac{dT}{dt} = -0.2t + 1.4
\][/tex]
2. Find Critical Points:
Critical points occur where the derivative is zero. We set the derivative equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[
-0.2t + 1.4 = 0
\][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[
t = \frac{1.4}{0.2} = 7
\][/tex]
3. Evaluate the Temperature at the Critical Point:
Next, we substitute this critical point back into the original function to find the corresponding temperature:
[tex]\[
T(7) = -0.1(7)^2 + 1.4(7) + 98.5
\][/tex]
Computing this gives:
[tex]\[
T(7) = -0.1(49) + 9.8 + 98.5 = -4.9 + 9.8 + 98.5 = 103.4
\][/tex]
4. Conclusion:
Therefore, at [tex]\( t = 7 \)[/tex] days, the temperature reaches a relative maximum of 103.4°F.
This process allows us to understand the temperature dynamics during this illness, revealing that the highest temperature occurs at day 7 with a value of 103.4°F.
1. Determine the Derivative:
The first step is to find the derivative of [tex]\( T(t) \)[/tex]. The derivative, [tex]\( \frac{dT}{dt} \)[/tex], gives the rate of change of temperature with respect to time and helps us find the critical points where the temperature might be at an extreme (either a maximum or a minimum).
The derivative of the function [tex]\( T(t) = -0.1t^2 + 1.4t + 98.5 \)[/tex] with respect to [tex]\( t \)[/tex] is:
[tex]\[
\frac{dT}{dt} = -0.2t + 1.4
\][/tex]
2. Find Critical Points:
Critical points occur where the derivative is zero. We set the derivative equal to zero and solve for [tex]\( t \)[/tex]:
[tex]\[
-0.2t + 1.4 = 0
\][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[
t = \frac{1.4}{0.2} = 7
\][/tex]
3. Evaluate the Temperature at the Critical Point:
Next, we substitute this critical point back into the original function to find the corresponding temperature:
[tex]\[
T(7) = -0.1(7)^2 + 1.4(7) + 98.5
\][/tex]
Computing this gives:
[tex]\[
T(7) = -0.1(49) + 9.8 + 98.5 = -4.9 + 9.8 + 98.5 = 103.4
\][/tex]
4. Conclusion:
Therefore, at [tex]\( t = 7 \)[/tex] days, the temperature reaches a relative maximum of 103.4°F.
This process allows us to understand the temperature dynamics during this illness, revealing that the highest temperature occurs at day 7 with a value of 103.4°F.
