Problem 33 - College Algebra

Question

Given the function \(g(x)=x^{2}+2 x\), simplify \(\frac{g(x)-g(a)}{x-a}, x \neq a\)

Step-by-Step Solution

Answer

The simplified expression is \( x + a + 2 \).

Step 1: Write the Original Expression

The given exercise is to simplify \( \frac{g(x) - g(a)}{x-a}, \ x eq a \). First, identify the function \( g(x) = x^2 + 2x \).

Step 2: Calculate \( g(a) \)

Substitute \( a \) into the function \( g \) to get \( g(a) = a^2 + 2a \).

Step 3: Substitute \( g(x) \) and \( g(a) \) into the Expression

Replace \( g(x) \) and \( g(a) \) in the expression: \( \frac{g(x) - g(a)}{x-a} = \frac{(x^2 + 2x) - (a^2 + 2a)}{x-a} \).

Step 4: Simplify the Numerator

Simplify the expression in the numerator: \((x^2 + 2x) - (a^2 + 2a) = x^2 + 2x - a^2 - 2a = (x^2 - a^2) + (2x - 2a) \).

Step 5: Factor the Numerator

Use the difference of squares to factor \( x^2 - a^2 \) as \((x-a)(x+a)\), and factor \(2(x-a)\) from the linear terms: \( (x-a)(x+a + 2) \).

Step 6: Simplify the Expression

Cancel \( (x-a) \) in the numerator and the denominator, leaving \( x + a + 2 \).

Step 7: Provide the Simplified Result

The simplified result is \( x + a + 2 \). This is the final expression, given that \( x eq a \).

Key Concepts

Difference quotient

A difference quotient is a fundamental concept in calculus. It's the backbone of how we understand rates of change, particularly in limits and derivatives. To calculate it, we need to differentiate between two function points. This exercise simplifies the expression \( \frac{g(x) - g(a)}{x-a} \), which is a typical example of a difference quotient. Here, \( g(x) \) is defined as a polynomial function, \( g(x) = x^2 + 2x \).
In this context, the difference quotient measures the average rate of change of \( g(x) \) over the interval from \( a \) to \( x \). Understanding difference quotients involves:

• Taking an initial function and evaluating it at two different points.
• Substituting these points into the difference quotient formula.
• Simplifying the resultant expression to gain insights into the function's behavior.

This quotient is critical when you learn about derivatives, as it approaches the derivative's definition when \( x \) approaches \( a \).

Algebraic simplification

Algebraic simplification involves reducing expressions in a way that retains their original value while making them easier to work with. In this problem, we start with the expression \( \frac{g(x) - g(a)}{x-a} \), calculating and substituting values to obtain a simplified form. To simplify effectively:

• First, compute \( g(a) \), the function's value at point \( a \), using the original formula \( g(x) = x^2 + 2x \).
• Substitute \( g(x) \) and \( g(a) \) into the expression.
• Focus on simplifying the numerator by combining like terms and applying algebraic identities, such as the difference of squares: \( x^2 - a^2 = (x-a)(x+a) \).
• Factorize wherever possible. Here, the linear terms \( 2x - 2a \) factor nicely with the common term \( x-a \).

After factoring common terms, cancel any similar terms in the numerator and denominator, resulting in a more straightforward expression. Simplification greatly assists in calculus, as it provides a clearer view of how functions behave and interact.

Polynomial functions

Polynomial functions are expressions consisting of variables raised to different powers, multiplied by coefficients. The function \( g(x) = x^2 + 2x \) given in this exercise is a polynomial of degree 2. Polynomial functions have distinct properties:

• They are continuous and smooth, with no breaks or sharp turns in their graphs.
• Their degree, defined by the highest power of the variable, determines the function's basic shape.
• Higher-degree polynomials can exhibit multiple turning points and inflection points, affecting their graph's curvature.

Understanding polynomial behavior is crucial for calculus. In the context of the given difference quotient, evaluating \( g(x) \) at different points allows us to explore the function's change over a specified interval. The specific structure of polynomial functions makes them convenient for both algebraic manipulation and calculus applications, such as finding limits and derivatives.