Problem 82 - 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. $$f(x)=\frac{1}{2} x^{2}+4 x$$

Step-by-Step Solution

Answer

The simplified difference quotient is \( x + 4 \).

Step 1: Write the Difference Quotient Formula

The difference quotient is given by the formula \( \frac{f(x+h) - f(x)}{h} \). We will use this formula to find the difference quotient for the function \( f(x) = \frac{1}{2}x^2 + 4x \).

Step 2: Substitute \( x+h \) into \( f(x) \) to Find \( f(x+h) \)

First, substitute \( x+h \) into the function. This gives us \( f(x+h) = \frac{1}{2}(x+h)^2 + 4(x+h) \).

Step 3: Expand \( f(x+h) \)

Expand the expression from the previous step: \( f(x+h) = \frac{1}{2}(x^2 + 2xh + h^2) + 4x + 4h = \frac{1}{2}x^2 + xh + \frac{1}{2}h^2 + 4x + 4h \).

Step 4: Calculate \( f(x+h) - f(x) \)

Now subtract \( f(x) = \frac{1}{2}x^2 + 4x \) from \( f(x+h) = \frac{1}{2}x^2 + xh + \frac{1}{2}h^2 + 4x + 4h \). This results in \( f(x+h) - f(x) = xh + \frac{1}{2}h^2 + 4h \).

Step 5: Divide by \( h \) to Get the Difference Quotient

Divide the expression from Step 4 by \( h \): \( \frac{xh + \frac{1}{2}h^2 + 4h}{h} = x + \frac{1}{2}h + 4 \).

Step 6: Simplify the Expression

Since we are taking the limit as \( h \) approaches zero in calculus, we can ignore terms that have \( h \) when simplifying. Thus, the simplified form of the difference quotient is \( x + 4 \).

Key Concepts

Understanding Algebraic Expressions

Algebraic expressions are combinations of numbers and letters linked by arithmetic operations such as addition, subtraction, multiplication, and division. They can include constants (like numbers) and variables (like letters) which represent unknown values or quantities that can change. For example, in the expression \( \frac{1}{2}x^2 + 4x \), \( x \) is a variable, and \( \frac{1}{2} \) is a coefficient that multiplies \( x^2 \). Also, the term \( 4x \) is another part of the expression where 4 is the coefficient of \( x \).

Algebraic expressions can be quite complex, involving factors, products, terms, and exponents. Being comfortable with these kinds of expressions is crucial because it allows you to model and solve real-world problems. Moreover, when working with functions as in this exercise, you need to substitute values into algebraic expressions accurately to find desired results. This skill is integral to evaluating functions and performing operations like the difference quotient.

Mastering Function Notation

Function notation is a way of representing a mathematical function using symbols. The most common form is \( f(x) \), where \( f \) denotes the function and \( x \) is the variable or input value. It tells you the rule to apply to \( x \) to get the output, usually referred to as \( f(x) \). For our problem, the function is given as \( f(x) = \frac{1}{2}x^2 + 4x \).

This notation allows us to evaluate the function at different points by substituting values for \( x \). For example, if we want to calculate \( f(x+h) \), we substitute \( x+h \) in place of \( x \) in our function. This is typical in difference quotient problems, where you find \( f(x+h) \) to understand how the function changes between \( x \) and \( x+h \).

Understanding function notation is essential for working with calculus concepts, and it provides a precise way to discuss and calculate changes in functions.

Simplifying Expressions Efficiently

Simplifying expressions involves reducing them to their simplest form without changing their value. This process makes expressions easier to work with in mathematical calculations and problem-solving.

To simplify the algebraic expression in a difference quotient like \( \frac{xh + \frac{1}{2}h^2 + 4h}{h} \), the process typically involves:

• Cancelling any common factors: Here, we can cancel \( h \) from each term in the numerator with \( h \) in the denominator.
• Combining like terms: This might not be applicable in every case, but it's an essential part of simplification.
• Eliminating unnecessary parts: In calculus, when finding limits, \( h \) approaches zero, allowing terms with \( h \) to be ignored for simplification.

Once simplified, the expression in our example becomes \( x + 4 \).

Simplifying expressions is a crucial skill in algebra and calculus because it makes the calculation of derivatives, integrals, and other operations much more manageable.