Question

Find the difference quotient of \(f\); that is, find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) for each function. Be sure to simplify. \(f(x)=\frac{2 x}{x+3}\)

Short answer

\(\frac{6}{(x+h+3)(x+3)}\)

Step-by-step solution

Substitute and Set up the Expression

Start by substituting the function into the difference quotient formula. the function is: \(f(x) = \frac{2x}{x+3}\)The difference quotient is given by: \[\frac{f(x+h) - f(x)}{h} = \frac{\frac{2(x+h)}{(x+h)+3} - \frac{2x}{x+3}}{h}eq 0\]

Find f(x+h)

First, find the expression for \( f(x+h) \). \[ f(x+h) = \frac{2(x+h)}{(x+h)+3} = \frac{2(x+h)}{x+h+3} \]

Combine the Fractions

Combine the fractions under a common denominator: \[\frac{ \frac{2(x+h)}{x+h+3} - \frac{2x}{x+3} }{ h } = \frac{ 2(x+h)(x+3) - 2x(x+h+3) }{ h (x+h+3)(x+3) } \]

Simplify the Numerator

Expand and simplify the numerator: \[2(x+h)(x+3) - 2x(x+h+3) = 2x^2 + 6x + 2hx + 6h - 2x^2 - 2hx - 6x = 6h\]So, the numerator simplifies to 6h.

Simplify the Entire Expression

Divide the entire expression by \( h \): \[\frac{ 6h }{ h (x+h+3)(x+3) } = \frac{ 6 }{ (x+h+3)(x+3) }\]

Key concepts

Difference Quotient

The difference quotient is a crucial concept in calculus that helps us understand the rate of change of a function. It is defined as \frac{ f(x+h) - f(x) }{ h }\ for any function f(x), where h is not equal to zero. The role of the difference quotient becomes clearer when you realize it is essentially the slope of the secant line connecting two points on the function. This idea lays the groundwork for the concept of the derivative.

To visualize, think of \( f(x) \) as a curve. The difference quotient finds the slope between any point \( x \) and another point \( x+h \) on this curve. As you let \( h \) approach zero, the secant line becomes the tangent line, giving us the derivative.

Simplifying Expressions

Simplifying expressions is a technique we use to make algebraic expressions more manageable. In the context of the difference quotient, it involves steps like combining fractions under a common denominator and canceling out terms. These steps are essential because they transform complex multi-step equations into simpler ones.

For instance, in our problem, simplifying the numerator allows us to cancel out terms and get to the final simplified form. By doing so, we can see that the numerator simplifies to \( 6h \), which makes it easier to work with the whole expression. Always keep in mind:

• Look for like terms to combine
• Factor where possible
• Simplify fractions by finding a common denominator

These steps help you manage and manipulate algebraic expressions effectively.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. Our function in this problem, \( f(x) = \frac{ 2x }{ x+3 } \), is a classic example. Understanding rational functions is key for solving difference quotients, as they often involve combining and simplifying these fractions.

Key points to remember about rational functions include:

• The domain excludes values that make the denominator zero.
• They have vertical asymptotes where the denominator equals zero.
• Horizontally, they approach a finite limit as \( x \) approaches infinity, depending on the degrees of the numerator and denominator.

In our example, the critical step is combining the given rational functions under a common denominator, making the simplification process straightforward.

Function Substitution

Function substitution is the process of replacing a variable with a given expression or another function. This technique is often used in the first step of solving difference quotients. In our problem, we started by substituting \( x+h \) into the function \( f(x) \).

Here’s a closer look at what we did:

• We replaced \( x \) with \( x+h \) in the function \( f(x) \).
• This gave us \( f(x+h) = \frac{2(x+h)}{x+h+3} \).
• Then we substituted both \( f(x+h) \) and \( f(x) \) into the difference quotient formula.

This substitution is critical as it sets up the foundation for further simplification. Remember, whenever you see \( f(x+h) \), you always start by substituting \( x+h \) into the function.