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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 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.

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.