Problem 56 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that a continuous function is periodic with period 2 and is quadratic between \(-0.25\) and \(0.25\) and linear between \(-1.75\) and \(-0.25 .\) In addition, it has the value 0 at 0 and \(0.0625\) at \(\pm 0.25 .\) Sketch the function over the domain \([-2,2]\), and give \(a\) piecewise definition of the function.

Step-by-Step Solution

Answer

The function is piecewise: quadratic on \([-0.25, 0.25]\) as \(x^2\), linear on \([-1.75, -0.25]\) as \(\frac{1}{24}x + 0.07292\), with a period of 2.

Step 1: Understand the Periodicity

The function is periodic with a period of 2. This means that if we define the function over the interval \([-1, 1]\), it will repeat the same pattern over the entire range \([-2, 2]\).

Step 2: Define the Quadratic Portion

The function is quadratic between \([-0.25, 0.25]\). We know the function has values 0 at 0 and 0.0625 at both -0.25 and 0.25. Assuming a simple form of \(f(x) = ax^2\) for this region, since the vertex (point at 0) of the quadratic would be the minimum, we set \(f(0.25) = 0.0625 = a(0.25)^2\), solving for \(a\), \(a = 1\). Therefore, the quadratic function is \(f(x) = x^2\).

Step 3: Define the Linear Portion

The function is linear between \([-1.75, -0.25]\). The line can be found using the two points \((-1.75, 0)\) and \(-0.25, 0.0625\). The slope \(m\) is calculated as \(m = \frac{0.0625 - 0}{-0.25 + 1.75} = \frac{0.0625}{1.5} = \frac{1}{24}\). The equation of the line is then \(y = mx + b\). Using the point (-0.25, 0.0625) to solve for \(b\), we have \(0.0625 = \frac{1}{24}(-0.25) + b\). Solving, \(b = 0.0625 + 0.01042 = 0.07292\). Thus, the linear function is \(f(x) = \frac{1}{24}x + 0.07292\).

Step 4: Define the Piecewise Function

From our findings, the function can be written as a piecewise function over one period \([-1, 1]\):\[f(x) = \begin{cases} x^2 & \text{if } -0.25 \leq x \leq 0.25 \\frac{1}{24}x + 0.07292 & \text{if } -1.75 < x < -0.25 \0 & \text{otherwise} \end{cases}\]

Step 5: Replicate the Periodic Function

Since the function has a period of 2, this piecewise form will need to be replicated over all periods. Thus over \([-2, 2]\), you replicate the pattern found over \([-1, 1]\) to cover \([-2, -1]\) and \([1, 2]\).

Key Concepts

Piecewise Function

A piecewise function is a function composed of multiple sub-functions, each applying to a specific interval in the domain. These intervals can vary in type, such as linear or quadratic, as seen in this exercise. Each segment of the piecewise function has its own formula or rule which dictates how its value is calculated.
For instance, in our exercise, the function is divided into two types of segments within specific intervals: quadratic from -0.25 to 0.25, and linear from -1.75 to -0.25. Outside these ranges, the function value is set to zero. This differentiation is essential as it allows the function to change its behavior and form in different domain sections, while maintaining continuity.
The piecewise nature allows us to effectively manage more complex functions by compartmentalizing them into simpler, more manageable sections, ensuring each part adheres to its own rules yet collectively forming a coherent function.

Quadratic Function

A quadratic function is a polynomial function of degree 2, taking the form of \(f(x) = ax^2 + bx + c\). The standard quadratic function is a parabola that opens upwards if the coefficient of the quadratic term \(a\) is positive.
In the given problem, we specifically consider the interval where the function is quadratic, that is between -0.25 and 0.25. Here, the quadratic function is \(f(x) = x^2\). This means there is no linear (\(b\)) or constant (\(c\)) term in the equation, simplifying the quadratic to a basic parabola with its vertex at the origin (point 0,0).
The quadratic section is thus symmetric about the y-axis because it contains no linear term, and the square term determines that it is always non-negative, matching the boundary conditions set in the problem, ensuring smoothness and continuity in the function structure.

Linear Function

A linear function is the simplest form of polynomial function, and takes the form \(f(x) = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
In our exercise, the linear function is relevant in the interval from -1.75 to -0.25. Within this range, the line is defined by points (-1.75, 0) and (-0.25, 0.0625), giving us a slope \(m = \frac{1}{24}\). This results in the linear equation \(f(x) = \frac{1}{24}x + 0.07292\).
The inclusion of such a function segment allows the overall function to maintain a simple, direct path of increase or decrease over a particular stretch of the domain. This linearity is key for managing the transition between different behavior intervals, offering a smooth passage from the quadratic segment to other portions of the function.

Function Sketching

Function sketching involves visualizing the behavior of the function across its domain. This requires plotting each piece of a piecewise function and ensuring continuity wherever segments meet.
In our exercise, we start by sketching the quadratic piece \(f(x) = x^2\) between -0.25 and 0.25, noting key points such as (0, 0) and (±0.25, 0.0625) which help define the parabola's curve. Next, the linear function \(f(x) = \frac{1}{24}x + 0.07292\) is plotted between -1.75 and -0.25, using its defining points to draw a straight line.
When sketching the entire function across the range \([-2, 2]\), it is vital to acknowledge its periodic nature. This means that the sketch drawn between \([-1, 1]\) must be repeated to fully capture the function's repetitive behavior across \([-2, -1]\) and \([1, 2]\). Sketching helps in verifying that all function conditions, continuity, and periodicity are maintained, offering a comprehensive view of its behavior across the entire domain.