Problem 90 - A Graphical Approach to College Algebra

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. $$f(x)=\frac{1}{x^{2}}$$

Step-by-Step Solution

Answer

The simplified difference quotient is \( \frac{-2x - h}{x^2(x + h)^2} \).

Step 1: Write the Given Function

We are given the function \( f(x) = \frac{1}{x^2} \). The task is to find the difference quotient \( \frac{f(x+h) - f(x)}{h} \).

Step 2: Evaluate f(x + h)

Substitute \( x + h \) into the function to find \( f(x + h) \): \[ f(x + h) = \frac{1}{(x + h)^2} \].

Step 3: Compute the Difference

Subtract \( f(x) \) from \( f(x+h) \): \[ f(x + h) - f(x) = \frac{1}{(x + h)^2} - \frac{1}{x^2} \].

Step 4: Find a Common Denominator

The expressions \( \frac{1}{(x+h)^2} \) and \( \frac{1}{x^2} \) have different denominators. The common denominator is \( x^2(x + h)^2 \). Rewrite each fraction:\[ \frac{1}{(x + h)^2} = \frac{x^2}{x^2(x + h)^2} \] \[ \frac{1}{x^2} = \frac{(x + h)^2}{x^2(x + h)^2} \].

Step 5: Simplify the Difference

Combine the fractions over the common denominator:\[ f(x + h) - f(x) = \frac{x^2 - (x + h)^2}{x^2(x + h)^2} \]Simplify the numerator:\[ x^2 - (x^2 + 2xh + h^2) = -2xh - h^2 \], so:\[ f(x + h) - f(x) = \frac{-2xh - h^2}{x^2(x + h)^2} \].

Step 6: Divide by h

Now, divide the entire expression by \( h \):\[ \frac{f(x + h) - f(x)}{h} = \frac{-2xh - h^2}{h \cdot x^2(x + h)^2} \].

Step 7: Simplify the Expression

Cancel \( h \) from numerator and denominator:\[ \frac{-2x - h}{x^2(x + h)^2} \].

Key Concepts

Functions

To understand the difference quotient, we first need to grasp the concept of functions. In mathematics, a function is a special relationship between two sets: a set of inputs (often called the domain) and a set of allowable outputs (the range). Every input is related to exactly one output. It's like a machine: you put something in, and you get something out.

For example, the function given in this exercise is \( f(x) = \frac{1}{x^2} \). This means that for every value of \( x \), there is a corresponding value of \( f(x) \). Here, \( f(x) \) tells us what happens when we input \( x \) into the equation \( \frac{1}{x^2} \), returning the reciprocal of the square of \( x \).

Understanding functions is crucial for solving the difference quotient as it helps us express how a function changes as its input changes.

Algebraic Fractions

Algebraic fractions are expressions that contain fractions with polynomials in the numerator and/or the denominator. They are similar to regular fractions, but instead of numbers, you have expressions involving variables, like \( \frac{1}{x^2} \).

In this exercise, we are dealing with algebraic fractions when working with \( f(x+h) = \frac{1}{(x + h)^2} \) and \( f(x) = \frac{1}{x^2} \). These expressions show us how the function behaves differently for different inputs.

• To perform operations like addition or subtraction with algebraic fractions, we need a common denominator.
• This involves finding the least common multiple of the denominators, so that the fractions can be combined into a single expression.

By mastering algebraic fractions, we can evaluate how a function changes between different values of \( x \), as seen in the solution where fractions were combined to find \( f(x + h) - f(x) \).

Simplification

Simplification is a key step in solving mathematical problems efficiently. It involves reducing expressions to their simplest form, making them easier to understand and work with.

For the difference quotient, simplification is crucial when combining and subtracting expressions. In the given solution, post obtaining the common denominator, the expression \( \frac{x^2 - (x + h)^2}{x^2(x + h)^2} \) is simplified to \( \frac{-2xh - h^2}{x^2(x + h)^2} \).

• This involves expanding any squares or other expressions in the numerator before simplifying.
• Post simplification, we cancel any common factors between the numerator and the denominator to further reduce the expression.

Simplification not only makes solving problems more manageable but also helps in revealing patterns or properties of functions more clearly.

Limits in Algebra

The concept of limits is a fundamental idea in calculus, which helps us understand the behavior of functions as they approach a specific point or infinity. In algebra, the difference quotient represents a foundational concept for finding a function's derivative using limits.

While calculating the difference quotient, understanding limits helps us make sense of expressions involving small changes, such as \( h \). As \( h \) approaches zero, the behavior of the expression \( \frac{f(x + h) - f(x)}{h} \) indicates the rate of change of the function at a particular point \( x \).

• The final simplified expression \( \frac{-2x - h}{x^2(x + h)^2} \) is crucial in finding the derivative using limit as \( h \to 0 \).
• This understanding aligns with finding instantaneous rates of change and further explores the concept of continuous functions.

Grasping limits allows you to transition smoothly from algebra to calculus, paving the way for exploring more complex topics.