Problem 38 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Evaluate \( f(a) \)

To find \( f(a) \), substitute \( a \) for \( x \) in the function. The function is given as \( f(x) = \frac{1}{x+1} \). Therefore: \[ f(a) = \frac{1}{a+1} \]

Step 2: Evaluate \( f(a+h) \)

Now, substitute \( a + h \) for \( x \) in the function. Thus, \[ f(a+h) = \frac{1}{(a+h)+1} = \frac{1}{a+h+1} \]

Step 3: Find the Difference Quotient

The difference quotient is given by the expression \( \frac{f(a+h) - f(a)}{h} \). Substitute the results from Step 1 and Step 2 into this expression:\[ \frac{\frac{1}{a+h+1} - \frac{1}{a+1}}{h} \]

Step 4: Simplify the Difference Quotient (Part 1)

Start by finding a common denominator for the terms in the numerator. The common denominator of \( \frac{1}{a+h+1} \) and \( \frac{1}{a+1} \) is \((a+h+1)(a+1)\):\[ \frac{1}{a+h+1} - \frac{1}{a+1} = \frac{a+1 - (a+h+1)}{(a+h+1)(a+1)} \]

Step 5: Simplify the Difference Quotient (Part 2)

Simplify the expression in the numerator:\[ a+1 - (a+h+1) = a+1 - a - h - 1 = -h \]So the expression becomes:\[ \frac{-h}{(a+h+1)(a+1)} \]

Step 6: Finalize the Difference Quotient

Complete the expression by dividing the entire simplified fraction by \( h \):\[ \frac{-h}{(a+h+1)(a+1)} \cdot \frac{1}{h} = \frac{-1}{(a+h+1)(a+1)} \]

Key Concepts

Function Evaluation

Function evaluation is the process of calculating the output of a function by substituting a specific input value into it. In this exercise, we are given the function \( f(x) = \frac{1}{x+1} \). When evaluating the function at \( x = a \), we need to replace \( x \) with \( a \). This step yields \( f(a) = \frac{1}{a+1} \). This is because the function notation \( f(x) \) indicates a relationship where each input \( x \) has a corresponding output. To further practice:

• Pick different values for \( a \), and plug them into the function to see what \( f(a) \) becomes.
• Try \( f(a+1) \) to understand how adding a constant to \( a \) affects the outcome.

Function evaluation is fundamental as it helps us understand how a function behaves for potential inputs.

Rational Functions

Rational functions are a type of function represented by the quotient of two polynomials. In our exercise, the rational function is \( f(x) = \frac{1}{x+1} \). This indicates that the numerator is a constant (1), while the denominator is a first-degree polynomial. Rational functions are intriguing because:

• They can exhibit complex behaviors, including vertical asymptotes and horizontal asymptotes.
• They can be undefined for certain values of \( x \), where the denominator equals zero, thus creating restrictions.

In the given function, it becomes undefined for \( x = -1 \) as substituting \( -1 \) in the denominator results in division by zero. Rational functions are vital in both mathematics and real-world applications because they can model relationships involving rates, proportions, or other comparative scenarios.

Common Denominator

The concept of finding a common denominator is crucial when dealing with the subtraction or addition of rational expressions. In the problem, we had to simplify the difference quotient \( \frac{f(a+h) - f(a)}{h} \). To perform the subtraction of the fractions \( \frac{1}{a+h+1} \) and \( \frac{1}{a+1} \), a common denominator was needed.Choosing the common denominator involves:

• Identifying the least common multiple of the denominators involved.
• The least common multiple for \( (a+h+1) \) and \( (a+1) \) is \((a+h+1)(a+1)\).

This common denominator provides a straightforward way to combine fractions by rewriting each original fraction:\[\frac{1}{a+h+1} - \frac{1}{a+1} = \frac{a+1 - (a+h+1)}{(a+h+1)(a+1)}\]Simplifying this, you achieve a rational expression that allows further manipulation of the difference quotient. Understanding common denominators allows you to handle operations between fractions more efficiently, a skill that is also useful in many areas of mathematics and calculations.