Problem 79 - 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. $$ f(x)=2 x^{2}-x+1 $$

Step-by-Step Solution

Answer

(a) \( f(x+h) = 2x^2 + 4xh + 2h^2 - x - h + 1 \); (b) Difference quotient: \( 4x + 2h - 1 \).

Step 1: Substitute for Function Evaluation

To find \( f(x + h) \), substitute \( x + h \) into the function \( f(x) = 2x^2 - x + 1 \). Replace every \( x \) with \( x + h \). This gives:\[ f(x + h) = 2(x + h)^2 - (x + h) + 1 \]

Step 2: Expand the Expression

Expand the squared term and distribute any remaining terms:1. \( (x + h)^2 = x^2 + 2xh + h^2 \)2. Therefore, \( 2(x + h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2 \)So, \[ f(x + h) = 2x^2 + 4xh + 2h^2 - x - h + 1 \]

Step 3: Find the Difference Quotient

The difference quotient is given by:\[ \frac{f(x+h) - f(x)}{h} \]Substitute \( f(x+h) \) with \( 2x^2 + 4xh + 2h^2 - x - h + 1 \) and \( f(x) \) with \( 2x^2 - x + 1 \). Substitute these into the equation:\[ \frac{(2x^2 + 4xh + 2h^2 - x - h + 1) - (2x^2 - x + 1)}{h} \]

Step 4: Simplify the Expression

Simplify the numerator:1. Cancel out the terms \( 2x^2 \), \( -x \), and \( +1 \) within the expression since they appear with opposite signs.2. The expression simplifies to: \( 4xh + 2h^2 - h \)Thus, the difference quotient simplifies to:\[ \frac{4xh + 2h^2 - h}{h} \]

Step 5: Final Simplification

Factor out \( h \) from the numerator:\[ \frac{h(4x + 2h - 1)}{h} \]Cancel the \( h \) in the numerator and denominator:\[ 4x + 2h - 1 \]This is the simplified form of the difference quotient.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables and coefficients. They are structured as sums of terms, each term being a product of a constant and a variable raised to an integer power. A basic polynomial function can be written as:

• \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\) - Here, \(a_n, a_{n-1}, \ldots, a_1, a_0\) are constants, also called coefficients.- The highest power (or exponent) of \(x\) in the function is known as the degree of the polynomial.

For example, in the polynomial function \(f(x) = 2x^2 - x + 1\),

• the degree is 2, - with coefficients 2, -1, and 1.

These types of functions are commonly used in algebra due to their straightforward structure and the variety of processes like differentiation and integration that can be easily applied to them. Understanding polynomial functions enables students to grasp various complex mathematical concepts through manageable pieces.

Function Evaluation

Function evaluation involves calculating the result of a function for a specific input value. In general terms, if you have a function \(f(x)\) and you need to find \(f(a)\), this simply means substituting the variable \(x\) with the value \(a\).

• For instance, to find \(f(x + h)\) for the function \(f(x) = 2x^2 - x + 1\), substitute \(x\) with \((x + h)\):- \(f(x + h) = 2(x + h)^2 - (x + h) + 1\).

This technique allows the determination of a new expression based on the function's formula. It is a fundamental skill used in calculus, particularly in calculating derivatives and understanding the behavior of polynomial functions across different values.

Algebraic Simplification

Algebraic simplification is the process of reducing expressions to their simplest form while maintaining equivalency. This involves operations like combining like terms, factoring, and canceling common factors. An important area where simplification is crucial is during the calculation of difference quotients.To simplify expressions,

• First, expand any squared terms, such as \((x + h)^2 = x^2 + 2xh + h^2\).
• Then, combine similar terms where possible to streamline the calculation.

For example, when finding the difference quotient \[ \frac{f(x+h) - f(x)}{h} \]for the function \(f(x) = 2x^2 - x + 1\),

• you simplify the expression to \(\frac{4xh + 2h^2 - h}{h}\),by canceling terms that negate each other across the numerator and denominator.

Finally, factor and reduce expressions like \(h(4x + 2h - 1)/h\), ensuring the cancellation of common factors. This results in a more concise and understandable expression, such as \(4x + 2h - 1\). Simplifying algebraic expressions is a key skill, laying the groundwork for more advanced mathematical problem-solving.