Problem 75 - A Graphical Approach to Precalculus with Limits

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) j for each function \(f\). Simplify completely. $$f(x)=4 x+3$$

Step-by-Step Solution

Answer

The difference quotient simplifies to 4.

Step 1: Substitute into the Difference Quotient Formula

The difference quotient formula is \( \frac{f(x+h)-f(x)}{h} \). Start by substituting \( f(x) = 4x + 3 \) into the formula: \( f(x+h) = 4(x+h) + 3 \).

Step 2: Expand the Expression for f(x+h)

Expand \( 4(x+h) + 3 \) to remove the parentheses. This becomes \( 4x + 4h + 3 \).

Step 3: Substitute and Simplify the Numerator

Substitute \( f(x+h) = 4x + 4h + 3 \) and \( f(x) = 4x + 3 \) into the difference quotient: \[ \frac{(4x + 4h + 3) - (4x + 3)}{h} \]. Simplify the expression by cancelling out terms: \[ \frac{4h}{h} \].

Step 4: Simplify the Fraction

Cancel \( h \) from the numerator and denominator: \( \frac{4h}{h} = 4 \). This is possible because \( h eq 0 \).

Key Concepts

Simplifying Expressions

Simplifying expressions is a crucial skill in mathematics, especially when dealing with algebraic formulas and equations. It involves reducing an expression to its simplest form. This process helps to clarify and solve mathematical problems more easily. To simplify an expression, follow these steps:

• Distribute any factors across terms within parentheses.
• Combine like terms, which are terms that have the same variables raised to the same power.
• Cancel out any common factors in the numerator and denominator if dealing with a fraction.

In contexts like the difference quotient, simplification is key. After finding the difference quotient for a given function, you'll often have extra terms that can be simplified further. This simplification helps in understanding the rate of change represented by the quotient. The final simplified form gives us a clearer picture of the behavior of functions, especially linear ones.

Linear Functions

Linear functions are among the simplest types of functions and are essential in precalculus. They are characterized by their constant rate of change. A linear function can be expressed in the form: \[ f(x) = mx + b \]where:

• \( m \) is the slope of the line, indicating the rate of change or steepness.
• \( b \) is the y-intercept, representing the point at which the line crosses the y-axis.

Linear functions graph as straight lines. When working with linear functions, the difference quotient becomes particularly straightforward. Since the change in function output divided by the change in input yields a constant slope, the simplified form of the difference quotient is the slope \( m \). This illustrates the predictable nature of linear functions, where the slope is consistent across the graph.

Precalculus

Precalculus serves as a bridge between algebra and calculus. It involves understanding concepts that form the foundation of calculus, such as functions, graphs, and limits. A key aspect of precalculus is the study of different types of functions, including linear functions, and their properties. The difference quotient is a fundamental concept in precalculus because it introduces the derivative concept seen in calculus. By working through difference quotients, students learn to conceptualize how functions change and how to determine the slope of a tangent line at any point. This understanding prepares students for more advanced studies in calculus, where they'll explore these rates of change in greater detail. Approaching precalculus with a firm grasp of simplifying expressions and understanding linear functions equips learners with the necessary tools for success in calculus. These concepts recur throughout mathematical education, making them integral to students' progression in mathematics.