Problem 39 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(f(a) = \frac{a}{a+1}\), \(f(a+h) = \frac{a+h}{a+h+1}\), difference quotient: \(\frac{1}{(a+h+1)(a+1)}\).

Step 1: Determine f(a)

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

Step 2: Determine f(a + h)

Next, find \(f(a+h)\) by substituting \(x = a + h\) into the function \(f(x) = \frac{x}{x+1}\). This yields: \[ f(a+h) = \frac{a+h}{a+h+1} \]

Step 3: Set up the Difference Quotient

The difference quotient is defined as \(\frac{f(a+h)-f(a)}{h}\). To set it up, we need: \[ f(a+h) - f(a) = \frac{a+h}{a+h+1} - \frac{a}{a+1} \]

Step 4: Simplify the Difference Quotient

To simplify \(\frac{f(a+h) - f(a)}{h}\), we first find a common denominator for \(f(a+h) - f(a)\). The common denominator is \((a+h+1)(a+1)\). Therefore: \[ f(a+h) - f(a) = \frac{(a+h)(a+1) - a(a+h+1)}{(a+h+1)(a+1)} \]This results in the numerator: \[ a^2 + ah + a + h - a^2 - ah - a = h \] Therefore, the expression simplifies to: \[ \frac{h}{(a+h+1)(a+1)} \]Finally, the difference quotient becomes: \[ \frac{f(a+h) - f(a)}{h} = \frac{1}{(a+h+1)(a+1)} \]

Key Concepts

Rational Function

A rational function is a type of function that is characterized by the ratio of two polynomials. In general, it can be expressed in the form \( f(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and the denominator \( Q(x) \) is not equal to zero. These functions are quite versatile and appear in various mathematical contexts.

In this particular exercise, the rational function given is \( f(x) = \frac{x}{x+1} \). This function involves the polynomial \( x \) in the numerator and \( x + 1 \) in the denominator. Remember that a key feature of rational functions is that the denominator cannot be zero, so the function is undefined at \( x = -1 \). This point will cause a vertical asymptote on the graph of the function.

Since rational functions form the basis of this exercise, understanding their properties is crucial for evaluating them and computing expressions like the difference quotient.

Function Evaluation

Function evaluation is the process of finding the value of a function for a specific input. It's like substituting a number into a mathematical expression to see what you get.

Let's break this down with the exercise. To find \( f(a) \), you replace every \( x \) in your function with \( a \). So it looks like this: \( f(a) = \frac{a}{a+1} \). This step is essential because knowing the value of \( f \) at a specific point helps us understand the behavior of the whole function.

Similarly, to find \( f(a+h) \), you substitute \( a + h \) in for \( x \), giving \( f(a+h) = \frac{a+h}{a+h+1} \). These substitutions are straightforward, yet they reveal how the function behaves as its input changes slightly.

• Evaluating functions at specific points helps in identifying trends or patterns.
• It is the first step when working towards calculating difference quotients.

This fundamental skill of substituting and simplifying will reappear in numerous mathematical problems.

Simplifying Expressions

Simplifying expressions is a critical mathematical skill, crucial when you want to make an expression more manageable to work with. In this exercise, the goal was to simplify the difference quotient \( \frac{f(a+h) - f(a)}{h} \).

Let's explore how to do this. First, find the common denominator for the two fractions \( \frac{a+h}{a+h+1} \) and \( \frac{a}{a+1} \), which is \((a+h+1)(a+1)\). Once you establish the common denominator, you can combine the fractions into a single expression.

Now, using the new common denominator, rewrite the expression:

• The numerator becomes \( (a+h)(a+1) - a(a+h+1) \).
• Simplify to the core expression by canceling and combining like terms: \( h \).

After simplification, the expression \( \frac{h}{(a+h+1)(a+1)} \) showcases how each component simplifies seamlessly.

This step-by-step simplification ensures that we accurately compute the symbolic representation of the difference quotient: \( \frac{1}{(a+h+1)(a+1)} \). Such reduction not only makes the expression cleaner but also offers a clearer insight into the function's behavior around the point \( a \).