Problem 76 - College Algebra with Modeling and Visualization

Question

Complete the following for the given \(f(x)\) (a) Find \(f(x+h)\) (b) Find the difference quotient of \(f\) and simplify. $$ f(x)=x^{2}-2 $$

Step-by-Step Solution

Answer

(a) \( f(x+h) = x^2 + 2xh + h^2 - 2 \); (b) The simplified difference quotient is \( 2x + h \).

Step 1: Find f(x+h)

To find \( f(x+h) \), substitute \( x+h \) into the function \( f(x) \) instead of \( x \). The original function is \( f(x) = x^2 - 2 \). Substituting gives: \[ f(x+h) = (x+h)^2 - 2 \] Simplify the expression: \[ (x+h)^2 = x^2 + 2xh + h^2 \] Therefore, \[ f(x+h) = x^2 + 2xh + h^2 - 2 \]

Step 2: Calculate the Difference Quotient

The difference quotient is given by the formula \( \frac{f(x+h) - f(x)}{h} \). Using the results from Step 1: \[ f(x+h) = x^2 + 2xh + h^2 - 2 \] and \[ f(x) = x^2 - 2 \], substitute these into the difference quotient: \[ \frac{f(x+h) - f(x)}{h} = \frac{(x^2 + 2xh + h^2 - 2) - (x^2 - 2)}{h} \] Simplify the expression: \[ \frac{x^2 + 2xh + h^2 - 2 - x^2 + 2}{h} \] which results in \[ \frac{2xh + h^2}{h} \].

Step 3: Simplify the Difference Quotient

Further simplify \( \frac{2xh + h^2}{h} \) by cancelling \( h \) from the numerator and the denominator: \[ 2x + h \] Hence, the simplified form of the difference quotient is \( 2x + h \).

Key Concepts

Polynomial Functions

Polynomial functions are foundational in algebra and calculus. They consist of terms composed of variables raised to whole number exponents, multiplied by coefficients. The function given in the exercise, \( f(x) = x^2 - 2 \), is a simple example of a polynomial function.

### Characteristics of Polynomial Functions

• **Degree**: This is determined by the highest exponent of the variable in the function. In \( x^2 - 2 \), the degree is 2, making it a quadratic polynomial.
• **Coefficients**: Numbers multiplying the terms. For \( x^2 - 2 \), the coefficient of \( x^2 \) is 1, and for the constant term, it is -2.
• **Terms**: Comprised of the variable and its exponent. The function has two terms: \( x^2 \) and -2.

Polynomial functions can be added, subtracted, and multiplied. When they are graphed, quadratic polynomials like \( x^2 - 2 \) form a parabola.

Understanding polynomials is essential in calculating derivatives, which involve finding the difference quotient as a step.

Algebraic Expressions

Algebraic expressions consist of numbers, variables, and operations like addition, subtraction, and multiplication. Expressions are the building blocks for forming polynomial functions.

### Manipulating Algebraic Expressions
To successfully evaluate algebraic expressions, especially when finding derivatives or integrals, one must be adept at:

• **Substitution**: This involves replacing variables with other expressions or values. For the given exercise, substituting \( x \) with \( x+h \) was crucial.
• **Simplification**: Reducing expressions to their simplest form, as done when \( (x+h)^2 \) was expanded to \( x^2 + 2xh + h^2 \).
• **Combining Like Terms**: Merging terms with the same variable and exponent, which aids simplification.

Learning to manipulate algebraic expressions is critical. It allows one to explore the nature of polynomial functions more profoundly and prepare for evaluating functions effectively.

Function Evaluation

Function evaluation is determining the output of a function given an input. It's a basic yet powerful concept that applies across math, particularly when working with polynomial functions.

### Steps in Evaluating Functions

• **Substitute Input**: Replace the variable in the function with the given input, like evaluating \( f(x) \) by finding \( f(x+h) \) in the given exercise.
• **Simplify Result**: Once the input is substituted, simplify the expression as much as possible to find the output value.
• **Interpret Output**: Understand the simplified result, which will either help solve a problem or gain insight into the function's behavior.

Evaluating functions is crucial for understanding how changes in the input affect the output. In calculus, it informs concepts like the difference quotient, leading to derivatives and analysis of function behavior. It's the gateway to exploring changes in functions, as shown when analyzing the polynomial \( f(x) = x^2 - 2 \) for different inputs.