Problem 86 - College Algebra with Modeling and Visualization

Question

What is the formula for the difference quotient? Given a formula for \(f(x),\) explain how to find \(f(x+h) .\) Give an example.

Step-by-Step Solution

Answer

The formula for the difference quotient is \( \frac{f(x+h) - f(x)}{h} \). For \( f(x) = x^2 \), \( f(x+h) = x^2 + 2xh + h^2 \).

Step 1: Understanding the Difference Quotient

The difference quotient is a formula used to calculate the average rate of change of a function over a small interval. It is given by the expression \( \frac{f(x+h) - f(x)}{h} \), where \( h \) represents a small increment in \( x \).

Step 2: Introducing the Function \(f(x)\)

Let us consider a specific function \( f(x) = x^2 \) as an example. Our goal is to apply the difference quotient formula to this function to find \( f(x+h) \) and subsequently calculate the average rate of change.

Step 3: Apply the Increment \(h\) to \(f(x)\)

To find \( f(x+h) \), substitute \( x+h \) into the function \( f(x) = x^2 \). This results in evaluating \((x+h)^2\).

Step 4: Expand the Binomial Expression

Expand \( (x+h)^2 \) using the binomial theorem: \( (x+h)^2 = x^2 + 2xh + h^2 \). This is the expression for \( f(x+h) \) when \( f(x) = x^2 \).

Step 5: Formulate the Difference Quotient

Using the expressions for \( f(x+h) = x^2 + 2xh + h^2 \) and \( f(x) = x^2 \), substitute them into the difference quotient formula: \( \frac{(x^2 + 2xh + h^2) - x^2}{h} \).

Step 6: Simplify the Expression

Simplify the expression by canceling the \( x^2 \) terms: \( \frac{2xh + h^2}{h} \). Further simplification yields \( 2x + h \). Thus, this represents the simplified difference quotient for this example.

Key Concepts

Average Rate of Change

The average rate of change of a function measures how much the function's value changes between two points on its graph, divided by the difference in the corresponding input values. This is similar to finding the slope of a line between two points, often called the secant slope.
For instance, if you have a function \( f(x) \) and you want to understand how it behaves from \( x \) to \( x+h \), you use the difference quotient formula:

• \( \frac{f(x+h) - f(x)}{h} \)

This expression essentially tells you how steep the function is over the interval \( h \). As \( h \) approaches zero, you get closer to knowing the instantaneous rate of change, which is a central concept in calculus. This leads to the derivative of the function.
Using our example function \( f(x) = x^2 \), if you're calculating between \( x \) and \( x+h \), you plug into the formula and find it simplifies to \( 2x + h \). As \( h \) becomes very small, \( 2x \) is the rate of change at any specific \( x \).

Function Evaluation

Function evaluation is the process of determining the output of a function for a given input. For a function defined as \( f(x) \), when you evaluate, you're simply substituting a specific numerical value of \( x \) into it to get the output. This is crucial in applying various mathematical concepts, like the difference quotient.
To evaluate a function such as \( f(x) = x^2 \), you might be asked to find \( f(x+h) \). What it means is that you replace \( x \) with \( x+h \) in the formula. This giving you:

• Substitute to find \( f(x+h) = (x+h)^2 \)
• This can be further calculated to get \( f(x+h) = x^2 + 2xh + h^2 \)

The evaluation of the function this way is essential when working with the difference quotient as it helps identify how the function's outputs change with small changes in inputs.

Binomial Theorem

The binomial theorem is a powerful algebraic tool used to expand expressions that are raised to a power, such as \((x + y)^n\). It's particularly helpful for calculating binomials in polynomial functions and is used extensively in calculus, particularly when evaluating difference quotients, as seen in the example.
For the binomial \((x+h)^2\), the theorem indicates that you can expand it as:

• \((x+h)^2 = x^2 + 2xh + h^2 \)

This expansion helps break down complex powers into manageable pieces that are easier to work with, especially when substituting into larger expressions. For example, in our situation with the function \( f(x) = x^2 \), you substitute \( x+h \) into the function, leading to the use of this outcome from the binomial theorem for a clear evaluation.
The expansion not only aids in calculations within the difference quotient but also helps in simplifying expressions and derivatives across calculus.