Question

If \(f(x)=L-x\) for \(0 < x < 2 L,\) and if \(f(x+2 L)=f(x),\) find a formula for \(f(x)\) in the interval \(-L < x < 0 .\)

Short answer

The formula for f(x) in the interval (-L, 0) is f(x) = x - L with x belonging to the interval (-L, 0).

Step-by-step solution

Use the periodic statement

We have the periodic statement \(f(x+2L)=f(x)\). Let's substitute the given function \(f(x)=L-x\): $$f(x+2L)=L-(x+2L)=L-x-2L=-x-L$$

Find a suitable value for x

We want to find a formula for \(f(x)\) in the interval \((-L,0)\). Let's set \(x=-L+y\), where \(y\) belongs to the interval \((0,L)\). Therefore, \(x\) belongs to the interval \((-L,0)\), which is the desired interval.

Substitute the value of x in the general formula

Now, let's substitute \(x=-L+y\) in the general formula \(f(x)=-x-L\): $$f(-L+y)=-(-L+y)-L=y-L$$

Final formula

Now we have the formula for \(f(x)\) in the interval \((-L,0)\): $$f(x)=x-L$$ with \(x\) belonging to the interval \((-L,0)\).

Key concepts

Periodicity in Functions

Imagine a heartbeat displayed on a monitor; it goes up and down in a continuous, rhythmic pattern. Just like that heartbeat, periodic functions have a regular pulse; they repeat their values at fixed intervals. In the world of mathematics, this 'heartbeat' is known as the period.

The period of a function is the smallest positive value for which the function's pattern repeats. For example, in the exercise, we are given that the function defined by \(f(x)=L-x\) for \(0 < x < 2L\) repeats itself every \(2L\) units. This means that \(f(x+2L)\) must be equal to \(f(x)\), defining the function's periodic nature. In essence, no matter how far you travel along the x-axis, if you move a distance of \(2L\), you will encounter the same function value.

Understanding the periodicity is critical when expanding the function across different intervals. In the step-by-step solution, when asked to find \(f(x)\) for \(-L < x < 0\), recognizing that the function's period is \(2L\) allows us to confidently assert that the function should behave identically in this alternate range, albeit with some necessary transformations.

Transformations of Functions

If the concept of periodic functions is the heartbeat, think of transformations as dance moves that can shift, stretch, or flip that heartbeat in various directions. Transformations of functions are like instructions on how to maneuver a graph around the coordinate plane without changing its shape.

Let's explore some common function dances, shall we?

• Translation: Think of it as a slide—moving a function up, down, left, or right. In our exercise, the function \(f(x)=L-x\) had to be translated to make sense in the interval \(-L, 0\).
• Reflection: This is akin to looking in a mirror. The function can be flipped over an axis. If we had \(f(x)=-x\), reflecting it over the y-axis would give us \(f(x)=x\).
• Stretching or Compression: This modifies how 'tall' or 'wide' our function is. If you multiply the function by a number greater than 1, it stretches; if you multiply it by a number between 0 and 1, it compresses.

In the provided exercise, by setting \(x=-L+y\), we perform a translation that moves the initially defined function into our target interval. This smart shift is needed to establish the function in the negative domain and helps us appreciate how transformations can unveil new expressions for the same periodic function in different intervals.

Piecewise-Defined Function

Life isn't always a smooth ride, and neither are functions—they can behave differently in different intervals. This is where piecewise-defined functions come into play; they allow a function to have different expressions depending on the part of the domain you're looking at. They are the 'choose-your-own-adventure' stories of the math world.

A piecewise-defined function is like a mathematical chameleon, changing its rule of operation depending on the 'environment' (which, in this case, is the interval of x-values). In the exercise, we see the hint of a piecewise-defined function where \(f(x)=L-x\) within the interval \(0 < x < 2L\), but the problem asks what happens for \(-L < x < 0\).

Utilizing the concept of periodicity and transformations, we've deduced a new piece of the function's rule. We end up with a two-part function:

• For \(0 < x < 2L\), we use \(f(x)=L-x\).
• And we've determined for \(-L < x < 0\), the function should be defined as \(f(x)=x-L\).

This is a simplified representation, but it illustrates the core of what piecewise functions are about. They piece together different behaviors to create a function that can effectively handle a diverse set of conditions.