Problem 82 - 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. $$f(x)=x^{2}+2 x$$

Step-by-Step Solution

Answer

The simplified difference quotient is \(2x + h + 2\).

Step 1: Write Down the Expression for f(x+h)

First, substitute \(x+h\) into the function \(f(x)=x^2+2x\). This will be \(f(x+h)=(x+h)^2+2(x+h)\).

Step 2: Expand f(x+h)

Now, expand the expression \((x+h)^2+2(x+h)\). - For \((x+h)^2\), use the formula \((a+b)^2 = a^2 + 2ab + b^2\) to get \(x^2 + 2xh + h^2\).- For \(2(x+h)\), distribute the 2 to get \(2x + 2h\).Combine these results to get \(x^2 + 2xh + h^2 + 2x + 2h\).

Step 3: Find f(x+h) - f(x)

Subtract \(f(x)=x^2+2x\) from your expanded \(f(x+h)=x^2 + 2xh + h^2 + 2x + 2h\). The \(x^2\) and \(2x\) terms will cancel out, leaving \(2xh + h^2 + 2h\).

Step 4: Form the Difference Quotient

The difference quotient is \(\frac{f(x+h)-f(x)}{h}\), which becomes \(\frac{2xh + h^2 + 2h}{h}\).

Step 5: Simplify the Difference Quotient

Factor out an \(h\) from the numerator: \(h(2x + h + 2)\). Cancel the \(h\) from numerator and denominator (since \(h eq 0\)), and you will get \(2x + h + 2\).

Key Concepts

Functions

A function is a fundamental concept in mathematics that relates elements from one set to another, typically using a rule or formula. The set of inputs is called the domain, and the possible outputs make up the range. In the function notation, like in the given exercise, \(f(x) = x^2 + 2x\), \(f\) denotes the function, \(x\) represents the input, and the expression \(x^2 + 2x\) is the rule that explains how to compute \(f(x)\).

• Functions can represent various real-world phenomena, from simple growth rates to complex systems.
• To evaluate a function for a certain input, substitute the input value into the function's expression.

Understanding functions is essential because they form the basis for more complex mathematical studies, including calculus, which is where the concept of a difference quotient comes from.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In the function \(f(x) = x^2 + 2x\), \(x^2 + 2x\) is an algebraic expression where:

• \(x^2\) represents the square of \(x\).
• \(+ 2x\) indicates two times the variable \(x\). This is known as a linear term.

Algebraic expressions can be manipulated by rules of algebra to simplify or transform them, which is essential in finding and simplifying the difference quotient.
The key to working with algebraic expressions lies in understanding operations like adding, subtracting, multiplying, and division. These operations allow us to express complex mathematical ideas in manageable formulas.

Simplification

Simplification is the process of making an algebraic expression as simple as possible. When dealing with the difference quotient, such as
\[\frac{f(x+h) - f(x)}{h}\] you must simplify the expression to make it manageable. Simplification typically involves:

• Cancelling terms when possible, as seen when subtracting \(f(x)\) from \(f(x+h)\).
• Combining like terms, such as collecting all terms involving \(h\) in the numerator.
• Factoring, as used in the expression \(h(2x + h + 2)\), to reduce terms.

Simplification not only eases calculations but is also crucial for ensuring computations are correct, especially when working with more complex functions. This step helps in identifying significant parts of the expression and discarding unnecessary complications.

Polynomial Functions

Polynomial functions are expressions that involve sums of powers of a variable. The function given, \(f(x) = x^2 + 2x\), is an example of a polynomial function.

• It is a second-degree polynomial because the highest power of the variable \(x\) is 2.
• Polynomial functions can have various terms, each consisting of a coefficient and a power of the variable. In \(f(x)\), the terms are \(x^2\) and \(2x\).

Polynomial functions are vital in mathematics due to their properties and the ability to approximate other functions.
Understanding how to manipulate these through operations like addition, subtraction, and multiplication plays a critical role in numerous mathematical processes, including calculating difference quotients.
Knowing the character of polynomial functions helps in predicting their behavior, which is useful not only in calculus but across various fields like physics and engineering.