Finding C Where F(c) Equals The Average Value A Calculus Problem

In this article, we delve into a fascinating problem in calculus: determining the value 'c' within a given interval where the function's value, f(c), matches the average value of the function over that interval. This concept elegantly combines the ideas of average value and function evaluation, providing a powerful tool for understanding function behavior. We will explore the underlying principles, walk through a step-by-step solution, and highlight the significance of this calculation in various applications.

Understanding the Problem

At its core, the problem asks us to find a specific point within an interval where the function's output coincides with its average output across the entire interval. The average value of a function over an interval represents the height of a rectangle with the same base (interval length) that has the same area as the area under the curve of the function within that interval. To find this average value, we use the following formula:

Average value = (1/(b-a)) ∫[a to b] f(x) dx

Where:

• f(x) is the function.
• [a, b] is the interval.
• ∫[a to b] f(x) dx represents the definite integral of f(x) from a to b, which calculates the area under the curve.

Once we determine the average value, the next step involves finding the 'c' value(s) within the interval [a, b] such that f(c) equals this average value. This typically involves solving an equation where we set the function f(x) equal to the calculated average value and then solve for x. The solutions within the interval [a, b] are the 'c' values we seek.

Step-by-Step Solution

Let's consider the specific problem presented: Given the function f(x) = x² + x - 6 on the interval [0, 4], find the value of c where f(c) equals the average value of the function over this interval. We'll break down the solution into manageable steps.

Step 1: Calculate the Average Value

First, we need to determine the average value of the function f(x) = x² + x - 6 on the interval [0, 4]. Using the formula mentioned earlier, we have:

Average value = (1/(4-0)) ∫[0 to 4] (x² + x - 6) dx

This requires us to evaluate the definite integral. Let's find the indefinite integral first:

∫(x² + x - 6) dx = (x³/3) + (x²/2) - 6x + C

Now, we apply the limits of integration (0 and 4):

∫[0 to 4] (x² + x - 6) dx = [(4³/3) + (4²/2) - 6(4)] - [(0³/3) + (0²/2) - 6(0)]

= (64/3) + 8 - 24 = (64/3) - 16 = (64 - 48)/3 = 16/3

Therefore, the average value is:

Average value = (1/4) * (16/3) = 4/3

Step 2: Set f(c) Equal to the Average Value

Now that we have the average value, which is 4/3, we need to find the value(s) of 'c' within the interval [0, 4] such that f(c) = 4/3. This means we need to solve the following equation:

c² + c - 6 = 4/3

To solve this quadratic equation, let's first get rid of the fraction by multiplying both sides by 3:

3(c² + c - 6) = 4

3c² + 3c - 18 = 4

Now, move all terms to one side to set the equation to zero:

3c² + 3c - 22 = 0

Step 3: Solve the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

c = [-b ± √(b² - 4ac)] / (2a)

Where a = 3, b = 3, and c = -22.

Plugging in the values, we get:

c = [-3 ± √(3² - 4 * 3 * -22)] / (2 * 3)

c = [-3 ± √(9 + 264)] / 6

c = [-3 ± √273] / 6

This gives us two possible solutions for 'c':

c₁ = (-3 + √273) / 6

c₂ = (-3 - √273) / 6

Step 4: Check if the Solutions are within the Interval

We need to ensure that our solutions lie within the interval [0, 4]. Let's approximate the values:

c₁ ≈ (-3 + 16.523) / 6 ≈ 13.523 / 6 ≈ 2.254

c₂ ≈ (-3 - 16.523) / 6 ≈ -19.523 / 6 ≈ -3.254

Since c₁ ≈ 2.254 falls within the interval [0, 4], it is a valid solution. However, c₂ ≈ -3.254 is outside the interval and is therefore not a valid solution.

Step 5: Round to the Nearest Thousandth

The problem asks us to round our answer to the nearest thousandth. Thus, the value of c where f(c) equals the average value is approximately:

c ≈ 2.254

Significance and Applications

Finding the value 'c' where f(c) equals the average value has significant implications in various fields:

• Physics: In physics, this concept can be used to determine the time at which the instantaneous velocity of an object equals its average velocity over a time interval. For example, consider an object moving with a velocity described by a function v(t). Finding the time 'c' where v(c) equals the average value of v(t) over an interval [a, b] tells us when the object's speed is momentarily equal to its overall average speed during that period. This is crucial in understanding the dynamics of motion and predicting future positions and velocities.

• Engineering: Engineers use this principle to analyze systems and optimize performance. For instance, in electrical engineering, it can help determine the point in time when the instantaneous power delivered to a circuit matches the average power consumption. Similarly, in mechanical engineering, it can be applied to find the point where the instantaneous stress on a material equals the average stress over a given loading cycle. This helps engineers design more efficient and reliable systems by ensuring components operate within safe and optimal ranges.

• Economics: In economics, this concept can be applied to analyze economic trends and predict future outcomes. For example, consider a function that models the inflation rate over time. Finding the time 'c' where the instantaneous inflation rate equals the average inflation rate over a specific period provides valuable insights into the stability of the economy and helps policymakers make informed decisions regarding monetary policy. It also aids in forecasting future inflation trends, allowing businesses and consumers to plan accordingly.

• Statistics: Statisticians use the concept of average value and its relationship to function values to analyze data sets and draw conclusions. For example, in quality control, the average value of a manufacturing process's output can be used to determine if the process is operating within acceptable limits. Finding the points where individual measurements deviate significantly from the average value helps identify potential issues and implement corrective actions. This ensures consistent product quality and minimizes production losses.

Conclusion

Determining the value of 'c' where f(c) equals the average value of a function is a fundamental concept in calculus with wide-ranging applications. It allows us to connect the instantaneous behavior of a function at a specific point with its overall behavior over an interval. By understanding the principles and applying the step-by-step solution, we can effectively solve these problems and gain valuable insights in various fields, from physics and engineering to economics and statistics. The process involves calculating the average value using definite integrals, setting the function equal to the average value, solving the resulting equation, and verifying that the solution lies within the given interval. This skill is crucial for anyone working with functions and seeking to understand their behavior in depth.

Keywords: average value, function, integral, interval, calculus, equation, physics, engineering, economics, statistics.