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Chapter 2: Problem 24

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

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{2x^2 - 4x + 3}{(2x - 3)^2} \); both the function and derivative are undefined at \( x = \frac{3}{2} \).

Step by step solution

01

Understand the Problem

We need to find the derivative of the function \( f(x) = \frac{x^2 - 1}{2x - 3} \) using the definition of the derivative. We also need to determine the domain of the function and its derivative.
02

Recall the Definition of the Derivative

The derivative of a function \( f(x) \) at a point \( x \) is given by the limit \( f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \). This definition will be used to find the derivative of \( f(x) \).
03

Compute \( f(x+h) \)

Substitute \( x+h \) into the function to get \( f(x+h) = \frac{(x+h)^2 - 1}{2(x+h) - 3} \). Simplifying, we get \( f(x+h) = \frac{x^2 + 2xh + h^2 - 1}{2x + 2h - 3} \).
04

Compute \( f(x+h) - f(x) \)

The expression becomes \( \frac{x^2 + 2xh + h^2 - 1}{2x + 2h - 3} - \frac{x^2 - 1}{2x - 3} \). To subtract these fractions, use a common denominator: \((2x + 2h - 3)(2x - 3)\).
05

Simplify the Difference

Perform the algebraic manipulation to combine the fractions. This involves expanding, simplifying, and then factoring terms where possible to cancel \( h \) after the subtraction is done.
06

Evaluate the Limit as \( h \to 0 \)

After simplification, evaluate \( \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \) which gives the derivative \( f'(x) \). After simplification, this yields \( f'(x) = \frac{2x^2 - 4x + 3}{(2x - 3)^2} \).
07

State the Domain of the Function

For the original function \( f(x) = \frac{x^2 - 1}{2x - 3} \), the domain is all real numbers except where the denominator is zero. Set \( 2x - 3 = 0 \) giving \( x = \frac{3}{2} \). Therefore, the domain is \( x \in \mathbb{R}, x eq \frac{3}{2} \).
08

State the Domain of the Derivative

The derivative \( f'(x) = \frac{2x^2 - 4x + 3}{(2x - 3)^2} \) is defined where the denominator is non-zero. Therefore, the domain of the derivative is also \( x \in \mathbb{R}, x eq \frac{3}{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a Function
The domain of a function refers to the set of all possible input values (x-values) for which the function is defined. This means the denominator cannot be zero, as division by zero is undefined. For the function \( f(x) = \frac{x^2 - 1}{2x - 3} \), the domain consists of all real numbers except the value of \( x \) that makes the denominator zero. To find this value, we set the denominator equal to zero:
  • \( 2x - 3 = 0 \)
  • Solve this equation for \( x \): \( 2x = 3 \Rightarrow x = \frac{3}{2} \)
Hence, the domain of \( f(x) \) is all real numbers except \( x = \frac{3}{2} \), or in mathematical notation, \( x \in \mathbb{R}, x eq \frac{3}{2} \). This ensures the function remains always defined.
Definition of Derivative
The derivative of a function is a fundamental concept in calculus. It measures how the function changes as its input changes. Formally, the derivative of a function \( f \) at a point \( x \) is defined as:\[ f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\]This expression represents the function's instantaneous rate of change at a particular point. In a geometric sense, it gives the slope of the tangent line to the curve at that point. When using this definition to find a derivative, we compute \( f(x+h) \) and simplify the expression. Then, we take the limit as \( h \) approaches zero. This is the definition we applied to calculate the derivative in the exercise.
Rational Functions
A rational function is a type of function that is the ratio of two polynomials. The general form of a rational function is \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The function \( f(x) = \frac{x^2 - 1}{2x - 3} \) is an example of a rational function. Key aspects of rational functions include:
  • They can have vertical asymptotes, which occur when the denominator is zero at some point.
  • The domain of a rational function excludes all x-values for which the denominator \( Q(x) \) is zero.
  • They can have holes, which occur where the numerator and denominator have common factors that are canceled out in simplified form.
Rational functions are central in understanding behavior and limits, especially when using derivatives to explore rates of change and other dynamic properties.

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