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Chapter 7: Problem 88

In each of Exercises \(85-88\), a function \(f\) is given. Let \(A_{f}(c)\) denote the average value of \(f\) over the interval \([c-1 / 4, c+1 / 4]\) Plot \(y=f(x)\) and \(y=A_{f}(x)\) for \(-1 \leq x \leq 1 .\) The resulting plot will illustrate the gain in smoothness that results from averaging. $$ f(x)=\left\\{\begin{array}{ccc} x^{2} & \text { if } & x<0 \\ x^{2}+x & \text { if } & 0 \leq x \end{array}\right. $$

Short Answer

Expert verified
Calculate the average value of the piecewise function over intervals and plot both the function and its averaged version to observe smoothing effects.

Step by step solution

01

Understanding the Function

The function \( f(x) \) is defined piecewise; it is \( x^2 \) for \( x < 0 \) and \( x^2 + x \) for \( x \geq 0 \). This means the function changes its formula at \( x = 0 \). We will use these definitions to calculate the average value over specified intervals.
02

Defining the Interval for Average Calculation

For a given \( c \), the interval is \([c - 1/4, c + 1/4]\). We'll find the average value of \( f \) over this interval by integrating \( f(x) \) over the interval and dividing by the interval's length, which is \(1/2\).
03

Setting Up the Integrals

Since the function is piecewise, we consider two main cases: when the interval doesn't include 0 and when it does. For ease, establish that the interval \([c - 1/4, c + 1/4]\) requires integration from two different sections if 0 is within the interval.
04

Case 1: Interval Entirely Less than Zero

If \( c < 1/4 \), then the entire interval \([c - 1/4, c + 1/4]\) lies to the left of 0, meaning we use \( f(x) = x^2 \). The average over the interval is calculated as: \[ A_f(c) = \frac{1}{1/2} \int_{c-1/4}^{c+1/4} x^2 \, dx \]
05

Case 2: Interval Entirely Greater than or Equal to Zero

If \( c > -1/4 \), then the entire interval \([c - 1/4, c + 1/4]\) lies to the right of 0 or starts at 0, meaning we use \( f(x) = x^2 + x \). Compute the average as:\[ A_f(c) = \frac{1}{1/2} \int_{c-1/4}^{c+1/4} (x^2 + x) \, dx \]
06

Case 3: Interval Contains Zero

If \( c = 0 \), split the integral at 0. The average value is the sum of integrals for \( x^2 \) and \( x^2 + x \) over their respective domains:\[ A_f(0) = \frac{1}{1/2} \left( \int_{-1/4}^{0} x^2 \, dx + \int_{0}^{1/4} (x^2 + x) \, dx \right) \]
07

Solving the Integrals

Calculate the definite integrals for each case:1. For \( f(x) = x^2 \), integrate and solve: \[ \int_{c-1/4}^{c+1/4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{c-1/4}^{c+1/4} \]2. For \( f(x) = x^2 + x \), solve as: \[ \int_{c-1/4}^{c+1/4} (x^2 + x) \, dx = \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{c-1/4}^{c+1/4} \]
08

Plotting the Functions

Plot original function \( y = f(x) \) along with \( y = A_f(x) \) over \(-1 \leq x \leq 1\) illustrating how averaging smooths out sharp transitions or changes in \( f \). Use a graphing tool if necessary to visualize this.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise Functions
Piecewise functions are fascinating as they allow us to define a single function in sections, each with its own formula. In this exercise, the function \( f(x) \) is defined differently for values less than and greater than zero. Specifically:
  • For \( x < 0 \), the function is \( f(x) = x^2 \).
  • For \( x \geq 0 \), it becomes \( f(x) = x^2 + x \).
What this means is that the behavior of the function depends on the interval we're looking at. It's like having two different rulebooks for different halves of the game.

Understanding how to work with piecewise functions is crucial because they are common in real-world scenarios where conditions change at certain thresholds.
Definite Integrals
The concept of definite integrals is foundational in calculus. It allows us to calculate the area under a curve over a specific interval. In our case, to find the average value of the function \( f(x) \), we need to evaluate definite integrals over the interval \([c - 1/4, c + 1/4]\).The integral measures the accumulation of the area under \( f(x) \) over a specified range. When computing for a piecewise function:
  • Case 1 requires integrating \( f(x) = x^2 \) entirely within the negative domain.
  • Case 2 calls for \( f(x) = x^2 + x \) over the positive domain.
  • Case 3 is unique because it splits the interval, handling the change at zero.
By calculating these integrals, we determine the total 'sum' of the function鈥檚 values over the interval and then divide by the interval length to find the average. This process gives us deeper insights into how the function behaves between different points.
Function Plotting
Plotting functions is a powerful way to visualize mathematical ideas. In this exercise, plotting both the original function \( y = f(x) \) and the averaged function \( y = A_f(x) \) over the range \(-1 \leq x \leq 1\) provides a visual understanding of how averaging affects the original curve.When you graph:
  • The original plot shows distinct changes or "corners," especially noticeable at \( x = 0 \).
  • The averaged function plot appears smoother as averaging "blends" values across intervals, reducing sharp turn points.
This visual representation helps students appreciate how the mathematical operations relate to geometric shapes. Graphical illustrations turn abstract calculus concepts into tangible forms.
Smoothness in Calculus
The notion of smoothness in calculus relates to how continuous and differentiable a function is. Smoothness often implies there are no abrupt holes, jumps, or sharp corners in the curve.By averaging a function over an interval, as we do with \( A_f(x) \), we achieve a degree of smoothing. This is because:
  • Averaging effectively "evens out" the peaks and troughs within a specific range, leading to a curve that does not have the steep or dramatic changes of the original.
  • This visual smoothness results in a function that is more regular and easier to predict or differentiate.
In calculus, smooth curves are generally easier to analyze and work with, especially when determining limits and derivatives as they provide consistent behavior across the interval.

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