Problem 22 - Calculus Early Transcendentals

Question

Use the fact that $$ \frac{d}{d x}[x \ln (2-x)]=\ln (2-x)-\frac{x}{2-x} $$ to show that the equation \(x=(2-x) \ln (2-x)\) has at least one solution in the interval \((0,1) .\)

Step-by-Step Solution

Answer

The function changes sign between 0 and 1, so there is at least one solution in (0,1).

Step 1: Understand the given equation

We start with the provided equation \(x = (2-x) \ln(2-x)\). We are asked to show that it has at least one solution in the interval \((0,1)\).

Step 2: Define a function based on the equation

Define a function \(f(x) = x - (2-x) \ln(2-x)\). Our goal is to show that this function has a root in the interval \((0,1)\), implying there is a point where \(f(x) = 0\).

Step 3: Evaluate the function at the endpoints of the interval

Calculate \(f(0)\) and \(f(1)\):\[ f(0) = 0 - (2-0)\ln(2-0) = -2 \ln 2 \]\[ f(1) = 1 - (2-1)\ln(2-1) = 1 - 0 = 1 \]

Step 4: Apply the Intermediate Value Theorem

Since \(f(0) = -2 \ln 2 < 0\) and \(f(1) = 1 > 0\), and \(f(x)\) is continuous (as both terms \(x\) and \((2-x)\ln(2-x)\) are continuous for \(x \in (0, 1)\)), by the Intermediate Value Theorem, there exists at least one \(c \in (0, 1)\) such that \(f(c) = 0\). This means that the equation \(x = (2-x) \ln(2-x)\) has at least one solution in this interval.

Key Concepts

Continuous Function

A continuous function is fundamental to many areas of calculus and provides the backbone for applying the Intermediate Value Theorem. In layman's terms, a function is continuous if you can draw its graph without lifting your pencil from the paper. This means there are no breaks, jumps, or holes in the graph of the function within the specific interval you are examining.

For a function to be regarded as continuous on an interval, it must meet the following criteria:

• The function is defined at every point in the interval.
• The limit of the function as it approaches any point within the interval exists.
• The value of the function at each point matches the limit at that point.

The function from the exercise, defined as \(f(x) = x - (2-x) \ln(2-x)\), is continuous over the interval \((0,1)\) because each component, \(x\) and \((2-x)\ln(2-x)\), are themselves continuous as long as \(x\) remains in \((0,2)\). This continuity is crucial for employing the Intermediate Value Theorem, ensuring there are no breaks where the root might be hidden.

Roots of Equations

When it comes to solving equations, the term 'roots' refers to the values of \(x\) that make the equation equal to zero. Finding the root of an equation is essentially solving the equation. For the equation given in the exercise, we are tasked with proving that there is at least one root in the interval \((0,1)\).

To demonstrate this, we turned the original equation \(x = (2-x) \ln(2-x)\) into a function \(f(x) = x - (2-x) \ln(2-x)\). The problem then boils down to finding \(x\) such that \(f(x)=0\), meaning we are looking for a point where the graph of the function crosses the x-axis.

The role of the Intermediate Value Theorem here is to confirm whether such a crossing happens between \(f(0)\) and \(f(1)\). Since \(f(0) < 0\) and \(f(1) > 0\), there must be at least one place where the function hits the x-axis, hence confirming the existence of a root.

Differentiation

Differentiation, the process of finding the derivative of a function, helps determine the rate of change of the function with respect to its variables. In our exercise, differentiation was used to confirm the given derivative \( \ln(2-x) - \frac{x}{2-x}\). Differentiation provides insights about the behavior of a function and is instrumental in checking function increases and decreases.

Let's talk about some basics of differentiation:

• The derivative of a function \( f(x) \) represents its slope at any point \( x \).
• If \( f'(x) > 0 \), the function is increasing at \( x \); if \( f'(x) < 0 \), the function is decreasing at \( x \).
• Higher derivatives can provide information about the concavity of the function.

For our specific purpose in the exercise, knowing the correct differentiation helped align the left-hand side of the equation with the right-hand side, enabling us to apply the Intermediate Value Theorem effectively.