Question

Simplify the difference quotient\(\frac{f(x+h)-f(x)}{h}\) for the following functions. $$f(x)=2 / x$$

Short answer

Question: Simplify the difference quotient for the function \(f(x) = \frac{2}{x}\). Answer: The simplified difference quotient for the given function is \(\frac{-2}{x(x+h)}\).

Step-by-step solution

Substitute f(x) and f(x+h) into the difference quotient formula

We have the difference quotient formula as: $$\frac{f(x+h)-f(x)}{h}$$ Now, let's substitute \(f(x)\) and \(f(x+h)\) for the given function \(f(x) = \frac{2}{x}\). We get: $$\frac{\frac{2}{x+h} - \frac{2}{x}}{h}$$

Simplify the numerator expression

To simplify the numerator, we need to find a common denominator for the two fractions: $$\frac{2}{x+h} - \frac{2}{x}$$ The common denominator for \(\frac{2}{x+h}\) and \(\frac{2}{x}\) is \((x+h) \cdot x\). Now, rewrite the fractions with the common denominator and simplify the expressions: $$\frac{2x - 2(x+h)}{(x+h)x}$$Expand the expression in the numerator and simplify:$$\frac{2x - 2x - 2h}{(x+h)x} = \frac{-2h}{(x+h)x}$$

Cancel out h term

Now we can cancel the \(h\) term from the numerator and the denominator: $$\frac{-2h}{(x+h)x \cdot h} \cdot \frac{1}{h} = \frac{-2}{x(x+h)}$$ The simplified difference quotient for the given function \(f(x) = \frac{2}{x}\) is: $$\frac{-2}{x(x+h)}$$

Key concepts

Simplifying Expressions

When simplifying expressions, especially those involving fractions and variables, the first step is often to consolidate similar terms and minimize the complexity of the expression. This can be achieved by combining like terms, factoring, and reducing fractions wherever possible.

For instance, when dealing with the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x) = \frac{2}{x}\), the simplification process starts by addressing the numerators of the complex fraction. The goal is to rewrite the complex fraction into a simpler form that is easier to work with. This often involves finding a common denominator, which allows us to combine the terms into a single fraction and then to cancel out any common factors between the numerator and the denominator.

Keeping the expressions simplified as much as possible at each step prevents errors and makes the subsequent steps clearer and easier to follow. As a result, the process not only becomes more efficient but also helps in developing a deeper understanding of algebraic manipulation and the properties of fractions.

Common Denominator

Finding a common denominator is crucial when dealing with fractions that need to be added, subtracted, or simplified. The common denominator is the least common multiple of the denominators of two or more fractions.

In the context of the difference quotient, the expression \(\frac{2}{x+h} - \frac{2}{x}\) requires a common denominator to combine the two fractions. This is because you cannot directly subtract fractions that have different denominators. The common denominator for \(\frac{2}{x+h}\) and \(\frac{2}{x}\) is the product of the two distinct denominators, \(x+h\) and \(x\), which results in \( (x+h) \cdot x \).

By rewriting each fraction with this common denominator, we can then subtract the fractions by subtracting their numerators, keeping the common denominator the same. This process simplifies the complex fraction to a single fraction with a common denominator, easing the path to further simplification such as cancellation of terms.

Rational Functions

Rational functions are expressions that can be written as the ratio of two polynomials. The difference quotient often leads to the simplification of a rational function, particularly when the function in question, like \(f(x) = \frac{2}{x}\), is itself a rational expression.

In the final stage of simplifying the difference quotient, having consolidated the fraction using the common denominator, we arrive at a new rational function \(\frac{-2h}{(x+h)x}\). The process doesn’t stop at finding a common denominator; it proceeds by identifying and cancelling common factors shared by the numerator and the denominator.

In this case, the factor \(h\) is present in both the numerator and the denominator and can be cancelled out. This cancellation is an important aspect of simplifying rational functions and requires a careful review to ensure that we only cancel factors, not terms that are being added or subtracted. After cancelling, the simplified difference quotient for \( f(x) \), which is itself a rational function, becomes \(\frac{-2}{x(x+h)}\). Understanding these types of functions is crucial for higher-level calculus concepts including limits, continuity, and asymptotic behavior.