Problem 40 - Algebra and Trigonometry

Question

Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0\) $$ f(x)=\frac{2 x}{x-1} $$

Step-by-Step Solution

Answer

The difference quotient is \( \frac{-2}{(a-1)(a+h-1)} \).

Step 1: Find f(a)

To find \(f(a)\), substitute \(a\) into the function \(f(x) = \frac{2x}{x-1}\). So, \(f(a) = \frac{2a}{a-1}\).

Step 2: Find f(a+h)

Substitute \(a+h\) for \(x\) in the function: \(f(a+h) = \frac{2(a+h)}{(a+h)-1} = \frac{2(a+h)}{a+h-1}\).

Step 3: Compute the Difference Quotient

The difference quotient is given by \(\frac{f(a+h)-f(a)}{h}\). So, substitute \(f(a+h)\) and \(f(a)\) from steps 1 and 2:\[f(a+h) - f(a) = \frac{2(a+h)}{a+h-1} - \frac{2a}{a-1}\]To compute this, find a common denominator and simplify:1. The common denominator is \((a-1)(a+h-1)\).2. Rewrite \( \frac{2(a+h)}{a+h-1} \) as \( \frac{2(a+h)(a-1)}{(a+h-1)(a-1)} \) and \( \frac{2a}{a-1} \) as \( \frac{2a(a+h-1)}{(a-1)(a+h-1)} \).3. Combine and simplify:\[\frac{2(a+h)(a-1) - 2a(a+h-1)}{(a-1)(a+h-1)}\]4. Simplify the numerator: - \(2(a+h)(a-1) = 2(a^2 + ah - a - h)\) - \(2a(a+h-1) = 2(a^2 + ah - a)\) - The difference: \(2(a^2 + ah - a - h) - 2(a^2 + ah - a) = -2h\)Thus: \( \frac{-2h}{(a-1)(a+h-1)} \)5. Divide by \(h\):\[\frac{f(a+h)-f(a)}{h} = \frac{-2h}{h(a-1)(a+h-1)} = \frac{-2}{(a-1)(a+h-1)}\]

Step 4: Final Expression: Difference Quotient

The difference quotient simplifies to \( \frac{-2}{(a-1)(a+h-1)} \). This is the finalized expression for the difference quotient.

Key Concepts

Function Evaluation

Evaluating a function means you're plugging in a specific value into the function equation to see what output it gives. Functions are like machines: you feed in one number, and out pops another number based on the function's rule. In this exercise, we start by finding \(f(a)\) for the function \(f(x) = \frac{2x}{x-1}\). This means replacing every \(x\) in the function with \(a\). Thus, \(f(a) = \frac{2a}{a-1}\).

Function evaluation doesn't change the structure of the function; it just provides a specific value. It is like pressing buttons on a calculator to get a result. To find \(f(a+h)\), you do the same process. Replace \(x\) with \(a+h\), which gives you \(f(a+h) = \frac{2(a+h)}{a+h-1}\). By evaluating a function at specific points, we can start to uncover patterns or rates of change, leading us to further insights such as the difference quotient.

Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. In this specific case, \(f(x) = \frac{2x}{x-1}\) is a rational function because it consists of two polynomial expressions, \(2x\) and \(x-1\). These kinds of functions feature prominently in algebra due to their interesting properties and behaviors such as asymptotes and undefined points.

It is critical to remember that rational functions can pose some challenges. For instance, they are undefined whenever the denominator equals zero. In this scenario, \(f(x)\) becomes undefined at \(x=1\), because it makes the denominator zero. Always consider the domain of the function—that is, the set of all permissible \(x\) values where the function is defined. Understanding these particulars can help in performing operations like calculating the difference quotient and simplifying expressions.

Simplifying Expressions

Simplification is an essential skill in mathematics. It makes complex expressions easier to work with and understand. In the context of finding the difference quotient, simplifying is crucial.

The difference quotient itself, \(\frac{f(a+h)-f(a)}{h}\), provides insight into the average rate of change of a function. To compute it, you first find \(f(a+h)\) and \(f(a)\) and then subtract them to get:

• \(f(a+h) - f(a) = \frac{2(a+h)}{a+h-1} - \frac{2a}{a-1}\)

To simplify this difference, you need a common denominator: \((a-1)(a+h-1)\). By rewriting the expressions with this common denominator, you can subtract them effectively.

• Numerator: Simplify by combining the terms.
• Denominator remains as the common denominator.

Once simplified, divide the numerator and denominator by \(h\). The expression is then reduced to \(\frac{-2}{(a-1)(a+h-1)}\), which represents the difference quotient simplification.

By systematically working through each step and simplifying, the results become clearer and more manageable, illustrating the power of simplification in mathematics.