Problem 56 - Calculus with Concepts in Calculus

Question

Let $$ T(x)=\left\\{\begin{array}{ll} 2 x & \text { for } 0 \leq x \leq \frac{1}{2} \\ 2(1-x) & \text { for } \frac{1}{2}

Step-by-Step Solution

Answer

The fixed point is \( x = \frac{2}{3} \).

Step 1: Consider the Definition of the Function

The definition of the piecewise function \( T(x) \) is given as follows:\[\T(x) = \begin{cases} 2x & \text{for } 0 \leq x \leq \frac{1}{2} \2(1-x) & \text{for } \frac{1}{2} < x \leq 1\end{cases}\]We want to find a value of \( x \) such that \( T(x) = x \), i.e., find \( x \) such that \( T(x) = x \).

Step 2: Solve for Fixed Point in the First Interval

First, solve for \( x \) in the interval \( 0 \leq x \leq \frac{1}{2} \).For this interval, \( T(x) = 2x \).Set up the equation:\[ 2x = x\]Subtract \( x \) from both sides:\[x = 0\]However, \( x = 0 \) is not in the open interval \((0, 1)\), so this value is not valid for our problem.

Step 3: Solve for Fixed Point in the Second Interval

Now, let's solve for \( x \) in the interval \( \frac{1}{2} < x \leq 1 \).For this interval, \( T(x) = 2(1-x) \).Set up the equation:\[ 2(1-x) = x\]Expand and simplify the equation:\[2 - 2x = x\]Add \( 2x \) to both sides:\[2 = 3x\]Divide both sides by 3:\[x = \frac{2}{3}\]Since \( \frac{2}{3} \) is in the interval \( \frac{1}{2} < x \leq 1 \), it is a valid solution.

Step 4: Verify the Solution

To verify, substitute \( x = \frac{2}{3} \) back into the function to check if \( T(x) = x \).\[ T\left(\frac{2}{3}\right) = 2\left(1 - \frac{2}{3}\right) = 2\left(\frac{1}{3}\right) = \frac{2}{3} \]Thus, \( T\left(\frac{2}{3}\right) = \frac{2}{3} \), confirming that \( x = \frac{2}{3} \) is indeed a fixed point for the function \( T(x) \).

Key Concepts

Piecewise Function

A piecewise function is a type of mathematical function defined by multiple sub-functions, each of which applies to a particular interval in the domain. Instead of having a single expression for all values of the independent variable, it "pieces" together different formulas over different parts of the domain. This allows for more complex and tailored behaviors, depending on the situation.
In the given problem, the function \( T(x) \) is piecewise. For \( 0 \leq x \leq \frac{1}{2} \), the function is defined as \( 2x \). For \( \frac{1}{2} < x \leq 1 \), it is defined as \( 2(1-x) \).
This approach lets the function model different scenarios based on the input \( x \). When working with piecewise functions, it's important to pay attention to the boundaries of each interval, as these dictate which formula to use.

• Identify: Notice the different sub-functions and the intervals they are defined over.
• Apply Appropriately: Always ensure you're using the right formula for the specified interval.

Understanding piecewise functions is crucial for solving problems that require different operations within specified ranges of the input variable. This is evident in exercises like this, where identifying and correctly applying the relevant part of the piecewise function is key to finding a solution.

Interval Notation

Interval notation is a way of describing a set of numbers along a number line. It is particularly helpful when expressing the domain or range of piecewise functions because it succinctly communicates where each sub-function is applicable. This notation can indicate whether endpoints are included or excluded in an interval.
In this exercise, the intervals \( 0 \leq x \leq \frac{1}{2} \) and \( \frac{1}{2} < x \leq 1 \) are specified using interval notation. The use of brackets \([\ ]\) and parentheses \((\ )\) indicates boundaries:

• Brackets \([\ ]\): Including an endpoint (e.g., \( \left[0, \frac{1}{2}\right] \) includes 0 and \( \frac{1}{2} \)).
• Parentheses \((\ )\): Excluding an endpoint (e.g., \( \left(\frac{1}{2}, 1\right] \) includes 1 but not \( \frac{1}{2} \)).

Using interval notation makes it clear which parts of the domain a sub-function applies to and ensures clarity when solving for fixed points as seen in this task. Recognizing these intervals helps in ensuring accurate application of each function part.

Function Verification

Function verification involves checking if a particular value satisfies the given function's conditions. This is important when you've found a potential solution, such as a fixed point, and want to confirm it's correct. In the context of the provided exercise, verifying the solution means confirming that when \( x = \frac{2}{3} \), the function \( T(x) \) indeed equals \( x \).
This verification involves substituting the value of \( x \) back into the equation to ensure it satisfies the original function. In this case:

• The condition \( T(x) = x \) means that the function evaluated at this point should return the point itself.

For \( T\left(\frac{2}{3}\right) \), we substitute into decided function part \( 2(1-x) \):
\[ T\left(\frac{2}{3}\right) = 2\left(1 - \frac{2}{3}\right) = 2\left(\frac{1}{3}\right) = \frac{2}{3} \]
Having \( T\left(\frac{2}{3}\right) = \frac{2}{3} \) confirms that this value is indeed a fixed point, proving our solution is accurate. Verification ensures the reliability of results, especially when dealing with piecewise functions.