Problem 59 - College Algebra

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$f(x)=x^{2}$$

Step-by-Step Solution

Answer

The simplified form of the difference quotient for the function \(f(x)=x^{2}\) is \(2x + h\).

Step 1: Substitute the function f into the difference quotient

Replace `f(x+h)` and `f(x)` in the difference quotient formula with the given function \(f(x)=x^{2}\). Thus the difference quotient becomes \(\frac{(x+h)^2 - x^2}{h}\).

Step 2: Expand the squared term

Expand the term \((x+h)^2\). Hence the difference quotient becomes \(\frac{x^2 + 2xh + h^2 - x^2}{h}\). The term \(x^2\) cancels out, leaving \(\frac{2xh + h^2}{h}\).

Step 3: Simplify the difference quotient

Finally, simplify this by dividing each term in the numerator by \(h\). This results in \(2x + h\), which is the simplified difference quotient.

Key Concepts

Simplifying Expressions

When working with algebra, simplifying expressions is an essential skill that allows students to break down complex problems into more manageable chunks. This process involves reducing the number of terms in an expression and combining like terms where possible.

For example, consider the difference quotient, which is used to find the rate at which a function's value is changing at a point. During the simplification of the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x)=x^2\), the expression \(x^2 + 2xh + h^2 - x^2\) appears. Here, the \(x^2\) terms are like terms and cancel each other out. Then, we are left with \(2xh + h^2\) in the numerator, which both contain the common factor \(h\).

By dividing each term by \(h\), we effectively reduce the complexity of the expression, yielding the simplified form \(2x + h\). It is crucial to recognize such opportunities to simplify, as they can make the remainder of a problem much easier to handle.

Function Notation

Function notation is a streamlined way of representing functions in algebra, allowing us to communicate and manipulate functions efficiently. In this notation, a function is usually denoted by \(f(x)\), where \(f\) is the name of the function and \(x\) is the variable input, often referred to as the 'argument' of the function.

In the context of the difference quotient, function notation plays a pivotal role. When the problem asks for \(\frac{f(x+h)-f(x)}{h}\), it involves evaluating the function \(f\) at two different inputs, \(x\) and \(x+h\). The notation makes it clear that you are dealing with the same function evaluated at two points along its domain, separated by an interval of \(h\).

This concept is fundamental to understanding calculus and rates of change, as it sets the stage for defining concepts like derivatives. It's also an expression of how mathematical operations can apply to functions, not just to numbers.

Algebraic Functions

Algebraic functions are mathematical expressions that involve variables and operations, such as addition, subtraction, multiplication, division, and exponentiation, like the function \(f(x) = x^2\) used in the difference quotient example.

These functions represent relationships between one or more variables. They are defined by algebraic expressions and can take various forms including linear, quadratic, polynomial, rational, and radical functions. The function \(f(x) = x^2\) is a quadratic function, characterized by the squared term \(x^2\), and represents a parabola when graphed.

Understanding the nature of algebraic functions, such as their shape, behavior, and the way they change, is critical for working with the difference quotient. When we apply the difference quotient to the quadratic function \(f(x) = x^2\), we can observe how the function's rate of change might vary depending on the value of \(x\). As foundations of algebra, grasping the properties and behaviors of these functions is key for success in more advanced mathematics.