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)