Problem 28 - Calculus: Early Transcendentals

Question

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( f(x) = \dfrac{x^2 - 1}{2x - 3} \)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{4x^2 - 9}{(2x - 3)^2} \) with domain \( \mathbb{R} \setminus \{3/2\} \).

Step 1: Define the Given Function and Identify Its Domain

The function given is \( f(x) = \frac{x^2 - 1}{2x - 3} \). The domain of this function is all real numbers except the value that makes the denominator zero. To find this, solve \( 2x - 3 = 0 \), which gives \( x = \frac{3}{2} \). Hence, the domain of \( f(x) \) is \( x \in \mathbb{R} \setminus \left\{ \frac{3}{2} \right\} \).

Step 2: Define the Derivative using the Limit Definition

The derivative of the function \( f(x) \) at any point \( x \) is defined by the limit: \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). We will use this definition to find \( f'(x) \).

Step 3: Substitute and Simplify the Expression

Substitute \( f(x+h) = \frac{(x+h)^2 - 1}{2(x+h) - 3} \) and \( f(x) = \frac{x^2 - 1}{2x - 3} \) into the limit definition. The expression becomes: \[ \lim_{h \to 0} \frac{\frac{(x+h)^2 - 1}{2(x+h) - 3} - \frac{x^2 - 1}{2x - 3}}{h}. \] Simplify the numerator to a single fraction: \[ \frac{((x+h)^2 - 1)(2x - 3) - (x^2 - 1)(2(x+h) - 3)}{(2(x+h) - 3)(2x - 3)}. \]

Step 4: Simplify the Expression Further

Expand the terms and simplify the expression: \((x^2 + 2hx + h^2 - 1)(2x - 3) - (x^2 - 1)(2x + 2h - 3) \). After simplification, this becomes a polynomial in terms of \( h \).

Step 5: Find the Limit as h Approaches Zero

Cancel out the \( h \) in the simplified fraction's numerator and take the limit as \( h \to 0 \). The terms containing \( h \) become zero, leaving a single expression for \( f'(x) \). Solve the limit to find that \[ f'(x) = \frac{(4x^2 - 9)}{(2x - 3)^2}. \]

Step 6: Determine the Domain of the Derivative

The domain of \( f'(x) \) is determined by the same restrictions as the original function. The denominator of \( f'(x) \), \((2x - 3)^2 \), must not be zero, thus \( x = \frac{3}{2} \) is excluded. Therefore, the domain of \( f'(x) \) is \( x \in \mathbb{R} \setminus \left\{ \frac{3}{2} \right\} \).

Key Concepts

Limit Definition of Derivative

The limit definition of a derivative is a fundamental concept in calculus. It provides a precise way to determine the rate at which a function changes at any point. For a given function \( f(x) \), the derivative \( f'(x) \) at a point \( x \) is defined as:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This definition relies on the idea of a slope of a tangent line to the curve at a single point. Imagine a tiny step \( h \) along the x-axis. The difference \( f(x+h) - f(x) \) represents the change in the function's value due to this step. The ratio of this change, divided by \( h \), shows how quickly the function is changing at \( x \). Then, by taking the limit as \( h \) approaches zero, we ascertain the instantaneous rate of change, i.e., the derivative. Understanding this concept helps you grasp the subtle yet powerful way calculus models real-world changes.

Domain of a Function

Understanding the domain of a function is crucial for defining where a function is valid and can be calculated. In the case of the function \( f(x) = \frac{x^2 - 1}{2x - 3} \), the primary concern is identifying values that cause the denominator to become zero. A function is undefined when its denominator is zero because division by zero is not possible. To find these critical values, solve the equation:\[2x - 3 = 0\]Solving gives \( x = \frac{3}{2} \). Therefore, the domain of this function consists of all real numbers except \( \frac{3}{2} \). In interval notation, it's denoted as \( x \in \mathbb{R} \setminus \left\{ \frac{3}{2} \right\} \). Similarly, the domain of the derivative \( f'(x) \) is also all real numbers except \( \frac{3}{2} \) since the denominator of \( f'(x) \) contains the factor \((2x - 3)^2\), reiterating that \( x = \frac{3}{2} \) needs to be excluded.

Polynomial Simplification

Polynomial simplification is a powerful skill in mathematics, enabling you to express complex expressions more clearly by combining like terms and reducing equations. When simplifying the difference quotient in the derivative of a rational function, you're often faced with polynomials. For instance, the expression:\[ (x^2 + 2hx + h^2 - 1)(2x - 3) - (x^2 - 1)(2x + 2h - 3)\]needs to be expanded and simplified. The goal is to reduce this into simpler terms, focusing on collecting like terms and canceling those that vanish when \( h \to 0 \).Key Steps:

• Expand each term involving \( h \) and \( x \).
• Combine like terms for each individual polynomial.
• Simplify by canceling terms, especially those involving \( h \) which will become insignificant as \( h \to 0 \).
• Factor and reduce final expressions for clarity.

By carrying out these steps, the complex polynomial is simplified, enabling easy calculation and interpretation of the original function's derivative.