Thank you for visiting Suppose that a person s heart rate tex x tex minutes after vigorous exercise has stopped can be modeled by the function tex f x. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Let's solve the problem step by step:
### (a) Evaluate [tex]\( f(0) \)[/tex] and [tex]\( f(4) \)[/tex]. Interpret the result.
The function provided is:
[tex]\[ f(x) = \frac{4}{5}(x-10)^2 + 71 \][/tex]
First, we need to evaluate [tex]\( f(0) \)[/tex]:
[tex]\[
f(0) = \frac{4}{5}(0-10)^2 + 71 = \frac{4}{5}(100) + 71 = 80 + 71 = 151
\][/tex]
Next, we evaluate [tex]\( f(4) \)[/tex]:
[tex]\[
f(4) = \frac{4}{5}(4-10)^2 + 71 = \frac{4}{5}(36) + 71 = 28.8 + 71 = 99.8
\][/tex]
Interpretation:
- [tex]\( f(0) = 151 \)[/tex]: The person's heart rate is 151 beats per minute initially, right after stopping exercise.
- [tex]\( f(4) = 99.8 \)[/tex]: The person's heart rate is about 99.8 beats per minute after 4 minutes.
Let's choose the correct interpretation:
A. The person's heart rate is 151 beats per minute initially and about 99.8 beats per minute after 4 minutes.
### (b) Estimate the times when the person's heart rate was between 100 and 120 beats per minute, inclusive.
To find these times, we need to solve the inequality:
[tex]\[ 100 \leq \frac{4}{5}(x-10)^2 + 71 \leq 120 \][/tex]
Solving for the lower bound (100 bpm):
[tex]\[ 100 = \frac{4}{5}(x-10)^2 + 71 \][/tex]
[tex]\[ \frac{4}{5}(x-10)^2 = 29 \][/tex]
[tex]\[ (x-10)^2 = 36.25 \][/tex]
Taking the square root:
[tex]\[ x-10 = \pm6.02 \][/tex]
[tex]\[ x = 10 \pm 6.02 \][/tex]
The potential solutions are approximately [tex]\( x = 3.98 \)[/tex] (within domain).
Solving for the upper bound (120 bpm):
[tex]\[ 120 = \frac{4}{5}(x-10)^2 + 71 \][/tex]
[tex]\[ \frac{4}{5}(x-10)^2 = 49 \][/tex]
[tex]\[ (x-10)^2 = 61.25 \][/tex]
Taking the square root:
[tex]\[ x-10 = \pm7.83 \][/tex]
[tex]\[ x = 10 \pm 7.83 \][/tex]
The potential solution is approximately [tex]\( x = 2.17 \)[/tex] (within domain).
Interpretation:
The heart rate was between 100 and 120 beats per minute from about 2.17 minutes to 3.98 minutes after exercise stopped.
So, the correct choice is:
B. About [tex]\( 2.2 \leq x \leq 4.0 \)[/tex] (min)
### (a) Evaluate [tex]\( f(0) \)[/tex] and [tex]\( f(4) \)[/tex]. Interpret the result.
The function provided is:
[tex]\[ f(x) = \frac{4}{5}(x-10)^2 + 71 \][/tex]
First, we need to evaluate [tex]\( f(0) \)[/tex]:
[tex]\[
f(0) = \frac{4}{5}(0-10)^2 + 71 = \frac{4}{5}(100) + 71 = 80 + 71 = 151
\][/tex]
Next, we evaluate [tex]\( f(4) \)[/tex]:
[tex]\[
f(4) = \frac{4}{5}(4-10)^2 + 71 = \frac{4}{5}(36) + 71 = 28.8 + 71 = 99.8
\][/tex]
Interpretation:
- [tex]\( f(0) = 151 \)[/tex]: The person's heart rate is 151 beats per minute initially, right after stopping exercise.
- [tex]\( f(4) = 99.8 \)[/tex]: The person's heart rate is about 99.8 beats per minute after 4 minutes.
Let's choose the correct interpretation:
A. The person's heart rate is 151 beats per minute initially and about 99.8 beats per minute after 4 minutes.
### (b) Estimate the times when the person's heart rate was between 100 and 120 beats per minute, inclusive.
To find these times, we need to solve the inequality:
[tex]\[ 100 \leq \frac{4}{5}(x-10)^2 + 71 \leq 120 \][/tex]
Solving for the lower bound (100 bpm):
[tex]\[ 100 = \frac{4}{5}(x-10)^2 + 71 \][/tex]
[tex]\[ \frac{4}{5}(x-10)^2 = 29 \][/tex]
[tex]\[ (x-10)^2 = 36.25 \][/tex]
Taking the square root:
[tex]\[ x-10 = \pm6.02 \][/tex]
[tex]\[ x = 10 \pm 6.02 \][/tex]
The potential solutions are approximately [tex]\( x = 3.98 \)[/tex] (within domain).
Solving for the upper bound (120 bpm):
[tex]\[ 120 = \frac{4}{5}(x-10)^2 + 71 \][/tex]
[tex]\[ \frac{4}{5}(x-10)^2 = 49 \][/tex]
[tex]\[ (x-10)^2 = 61.25 \][/tex]
Taking the square root:
[tex]\[ x-10 = \pm7.83 \][/tex]
[tex]\[ x = 10 \pm 7.83 \][/tex]
The potential solution is approximately [tex]\( x = 2.17 \)[/tex] (within domain).
Interpretation:
The heart rate was between 100 and 120 beats per minute from about 2.17 minutes to 3.98 minutes after exercise stopped.
So, the correct choice is:
B. About [tex]\( 2.2 \leq x \leq 4.0 \)[/tex] (min)
Thank you for reading the article Suppose that a person s heart rate tex x tex minutes after vigorous exercise has stopped can be modeled by the function tex f x. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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