Problem 32 - 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

\(f(a)=\frac{1}{a+1}, f(a+h)=\frac{1}{a+h+1},\) and the difference quotient is \(\frac{-1}{(a+1)(a+h+1)}.\)

Step 1: Find f(a)

To find \( f(a) \), substitute \( x = a \) into the function \( f(x) = \frac{1}{x+1} \). This gives: \[ f(a) = \frac{1}{a+1}. \]

Step 2: Find f(a+h)

To find \( f(a+h) \), substitute \( x = a + h \) into the function \( f(x) = \frac{1}{x+1} \). This gives: \[ f(a+h) = \frac{1}{(a+h)+1} = \frac{1}{a+h+1}. \]

Step 3: Calculate the Difference Quotient

The difference quotient is given by the formula \( \frac{f(a+h)-f(a)}{h} \). Substituting \( f(a+h) = \frac{1}{a+h+1} \) and \( f(a) = \frac{1}{a+1} \), we have: \[ \frac{f(a+h)-f(a)}{h} = \frac{\frac{1}{a+h+1} - \frac{1}{a+1}}{h}. \] To simplify the difference, use the common denominator \((a+1)(a+h+1)\): \[ \frac{1}{a+h+1} - \frac{1}{a+1} = \frac{(a+1) - (a+h+1)}{(a+1)(a+h+1)}. \] The numerator simplifies to \( a+1 - a -h - 1 = - h \), so: \[ \frac{-h}{(a+1)(a+h+1)}. \] The difference quotient becomes: \[ \frac{-h}{h(a+1)(a+h+1)} = \frac{-1}{(a+1)(a+h+1)}. \]

Key Concepts

Rational Functions

Rational functions are mathematical expressions that involve the ratio of two polynomials. Essentially, these are fractions where the numerator and the denominator are polynomials. The function given in the exercise, \( f(x)=\frac{1}{x+1} \), is a quintessential example of a rational function. Here, the numerator is constant, and the denominator is a basic linear polynomial, \( x+1 \).

Rational functions are characterized by their vertical and horizontal asymptotes, which are lines that the graph of the function approaches but never actually touches. In this case, the rational function has a vertical asymptote at \( x = -1 \) because the function is undefined there. This happens because the denominator becomes zero, making the function undefined.

Being familiar with rational functions helps in graphing and understanding the behaviors of these functions as they approach different values. It’s important to note where the functions are not defined and how the functions behave as \( x \) approaches positive or negative infinity.

Function Evaluation

Function evaluation is the process of finding the value of a function for a specific input. This involves substituting a given value into the function in place of the variable and then performing the corresponding calculations.

In this exercise, we perform function evaluation when we find \( f(a) \) and \( f(a+h) \). For \( f(a) \), we substitute \( a \) into the function: \( f(a) = \frac{1}{a+1} \). Similarly, for \( f(a+h) \), we substitute \( a+h \): \( f(a+h) = \frac{1}{a+h+1} \).

Understanding the process of function evaluation is critical, as it allows us to compute values for specific inputs, which is particularly useful in calculus and other branches of mathematics that depend on analyzing the behavior of functions at specific points.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra that involves rewriting equations or expressions in a more concise and interpretable form. This often involves combining like terms, reducing fractions to their simplest terms, and factoring where possible.

In the context of the difference quotient in the exercise, simplifying involved a few key steps. First, notice that the expression, \( \frac{f(a+h) - f(a)}{h} \), involves a difference of fractions. The approach to simplify this was to find a common denominator, here \((a+1)(a+h+1)\), and then express the difference in terms of this denominator.

Once the difference of fractions is written with a common denominator, the next step is to simplify the numerator. In the exercise, this led to an expression \( -h \) in the numerator. Finally, simplifying the entire expression allowed us to cancel out \( h \) from the numerator and denominator, leading to the final, simplified form of the difference quotient: \( \frac{-1}{(a+1)(a+h+1)} \).

This process helps in reducing complexity and makes it easier to handle and interpret mathematical expressions, particularly in calculus.