Problem 71 - Brief Applied Calculus

Question

$$ \begin{array}{l} \text { For each function, find and simplify }\\\ \frac{f(x+h)-f(x)}{h} . \quad(\text { Assume } h \neq 0 .) \end{array} $$ $$ f(x)=\frac{2}{x} $$

Step-by-Step Solution

Answer

The simplified form is \(\frac{-2}{x(x+h)}\).

Step 1: Substitute into the Difference Quotient Formula

Start by substituting the function \(f(x) = \frac{2}{x}\) into the difference quotient formula \(\frac{f(x+h)-f(x)}{h}\). This gives us \(\frac{\frac{2}{x+h} - \frac{2}{x}}{h}\).

Step 2: Find a Common Denominator

To subtract the fractions \(\frac{2}{x+h}\) and \(\frac{2}{x}\), find a common denominator, which is \(x(x+h)\). Rewrite each fraction with this common denominator: \(\frac{2x - 2(x+h)}{x(x+h)}\).

Step 3: Simplify the Numerator

Expand and simplify the numerator of the fraction: \(2x - 2(x+h) = 2x - 2x - 2h = -2h\). This gives us the expression \(\frac{-2h}{x(x+h)}\).

Step 4: Divide by \(h\)

Substitute \(-2h\) back into the difference quotient expression, resulting in \(\frac{-2h}{hx(x+h)}\) and simplify by canceling \(h\). This gives \(\frac{-2}{x(x+h)}\).

Step 5: Simplify the Expression

The expression \(\frac{-2}{x(x+h)}\) is already simplified. Thus, the function’s difference quotient is found.

Key Concepts

Rational Functions

Rational functions are a critical concept in mathematics, particularly in calculus. They are defined as functions expressed as the ratio of two polynomials. In simple terms, a function is rational if you can write it as \( \frac{p(x)}{q(x)} \), where both \( p(x) \) and \( q(x) \) are polynomials and \( q(x) eq 0 \).
Rational functions can exhibit features like vertical asymptotes, horizontal asymptotes, and potentially holes in the graph depending on the polynomials used.

• Vertical Asymptotes: Occur where the denominator is zero, indicating points where the function doesn't exist.
• Horizontal Asymptotes: Describe the behavior as \( x \) approaches infinity.

For the function \( f(x) = \frac{2}{x} \), it's a simple rational function since the numerator is a constant (2), and the denominator is a simple linear polynomial \( x \). Understanding this form helps when working through calculus problems involving limits and derivatives, as these fundamental properties will predict the function's behavior.

Simplifying Expressions

Simplifying expressions, especially when dealing with fractions like in rational functions, is about making them as compact and understandable as possible. When tackling a difference quotient, such as \( \frac{f(x+h)-f(x)}{h} \), steps to simplify require finding common denominators and reducing complex expressions.

To subtract \( \frac{2}{x+h} \) from \( \frac{2}{x} \), we need a common denominator. In this exercise, it was \( x(x+h) \). Once rewritten, fraction subtraction becomes straightforward, and terms in the numerator can be expanded and simplified.

• Common Denominator: Essential for subtracting terms, ensuring calculations are accurate.
• Canceling Terms: Once simplified, unnecessary terms (like \( h \) in this exercise) can often be canceled from numerators and denominators simplifying further.

Mastering these steps is crucial, as they form the backbone of algebraic manipulation needed in more advanced calculus problems.

Calculus Fundamentals

Calculus provides the tools to study change and motion, with the difference quotient being a vital early concept. This quotient enables us to understand how a function changes between two points and leads us to the derivative.

The difference quotient takes on the form \( \frac{f(x+h) - f(x)}{h} \), aiming to calculate the slope of the secant line between two points on the graph of a function. When the variable \( h \) approaches zero, this average rate of change becomes instantaneous, which is the foundational concept of derivatives.

• Secant Line: Connects two points on a curve, giving an average rate of change.
• Derivative: As \( h \to 0 \), the secant line approaches the tangent line, providing the exact rate of change at a point.

Grasping these concepts is essential as they underpin the study of calculus. They allow deeper insights into how functions behave, making it easier to solve complex real-world and theoretical problems.