Problem 131 - A Graphical Approach to College Algebra

Question

Solve each problem. Suppose that a person's heart rate, \(x\) minutes after vigorous exercise has stopped, can be modeled by $$f(x)=\frac{4}{5}(x-10)^{2}+80$$ The output is in beats per minute, where the domain of \(f\) is \(0 \leq x \leq 10\) (a) Evaluate \(f(0)\) and \(f(2) .\) Interpret the result. (b) Estimate the times when the person's heart rate was between 100 and 120 beats per minute, inclusive.

Step-by-Step Solution

Answer

f(0)=160, f(2)=131.2; heart rate is between 100-120 bpm from 2.93 to 5 minutes.

Step 1: Evaluate Function at a Specific Point

To find \( f(0) \), substitute \( x = 0 \) into the function:\[f(0) = \frac{4}{5}(0-10)^{2} + 80 = \frac{4}{5} \times 100 + 80 = 80 + 80 = 160\]Thus, the heart rate at \( x = 0 \) minutes is 160 beats per minute.

Step 2: Evaluate Function at Another Specific Point

To find \( f(2) \), substitute \( x = 2 \) into the function:\[f(2) = \frac{4}{5}(2-10)^{2} + 80 = \frac{4}{5} \times 64 + 80 = \frac{256}{5} + 80 = 51.2 + 80 = 131.2\]Thus, the heart rate at \( x = 2 \) minutes is approximately 131.2 beats per minute.

Step 3: Interpret the Results

Thus, \( f(0) = 160 \) means the heart rate right after stopping the exercise is 160 beats per minute. Similarly, \( f(2) = 131.2 \) indicates that after 2 minutes, the heart rate decreases to approximately 131.2 beats per minute.

Step 4: Estimate Time for Heart Rate Between 100 and 120

To find when \( 100 \leq f(x) \leq 120 \), solve the inequality:\[100 \leq \frac{4}{5}(x-10)^2 + 80 \leq 120\]Simplify and solve for \( x \):\[20 \leq \frac{4}{5}(x-10)^2 \leq 40\]\[\frac{5}{4} \times 20 \leq (x-10)^2 \leq \frac{5}{4} \times 40\]\[25 \leq (x-10)^2 \leq 50\]Taking the square root gives:\[5 \leq |x-10| \leq \sqrt{50}\]So, \( 5 \leq 10-x \leq \approx 7.07 \).By analyzing the situation, \( 10-7.07 \leq x \leq 10-5 \), which simplifies to \( 2.93 \leq x \leq 5 \).Hence, the heart rate is between 100 and 120 beats per minute from approximately 2.93 minutes to 5 minutes after stopping exercise.

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that has the general form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. In our example, the heart rate function is modeled using the quadratic function:

• \( f(x) = \frac{4}{5}(x-10)^2 + 80 \)
• This represents a parabola that opens upward because the term \( (x-10)^2 \) is positive.
• The function's minimum value may indicate the lowest heart rate after exercise.

Quadratic functions are very useful in modeling real-life scenarios where relationships are not linear, such as the decline in heart rate after exercise. Here, the vertex form \( \frac{4}{5}(x-10)^2 + 80 \) allows us to easily determine the vertex of the parabola which is a point at its lowest value.

Domain and Range

Understanding the domain and range of a function is crucial in determining where the function is applicable and what outputs it can produce.

The domain is the set of all possible input values (\( x \)) for the function. In our exercise, the domain of the heart rate function is \( 0 \leq x \leq 10 \). This means the function models the heart rate from the moment exercise stops until 10 minutes after.

Meanwhile, the range is the set of all possible output values (\( f(x) \)). Evaluations like \( f(0) = 160 \) indicate that starting heart rates can be quite high. As the exercise noted, the function decreases as \( x \) increases, reaching lower values, likely reaching a minimum around \( x = 10 \). This makes the range approximately from 80 beats per minute (the base addition) up to the maximum rate evaluated at the start.

Inequality Solving

Inequality solving often involves finding the range of \( x \) values where a condition is true. In our problem, we are looking for times when the heart rate is between 100 and 120 beats per minute.

Given:

• \( 100 \leq \frac{4}{5}(x-10)^2 + 80 \leq 120 \)
• We simplify this by subtracting 80: \( 20 \leq \frac{4}{5}(x-10)^2 \leq 40 \).
• To isolate \((x-10)^2\), multiply by \( \frac{5}{4} \) resulting in: \( 25 \leq (x-10)^2 \leq 50 \)

The next step involves taking square roots, leading to \( 5 \leq |x-10| \leq \sqrt{50} \). Understanding absolute values means we consider two cases and solve for \( x \). The solution involves manipulating ranges, giving us \( 2.93 \leq x \leq 5 \) minutes.

Exercise Physiology

Exercise physiology explores how physical activity changes body functions and structures. One key focus is how the cardiovascular system responds to exercise.

When exercising, heart rate increases to supply muscles with more oxygen and nutrients. After stopping, it gradually declines back to resting levels, often analyzed to understand cardiovascular health and fitness levels.

In this example:

• We use a quadratic function to model heart rate deceleration.
• Heart rate monitoring helps ensure safe exercise levels.
• Understanding recovery times aids in optimizing training and identifying potential health issues.

These physiological insights help in designing effective exercise programs, ensuring balance between exercise intensity and recovery periods.