Problem 107 - Precalculus : Building Concepts and Connections

Question

In Exercises \(105-110\), find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) \(h \neq 0,\) for the given function \(f\). $$f(x)=-x^{2}+x$$

Step-by-Step Solution

Answer

The difference quotient of the function \(f(x) = -x^{2} + x\) is \(-2x - h + 1\).

Step 1: Substitute \(x + h\) into \(f(x)\)

Begin with replacing the \(x\) in the function \(f(x) = -x^{2} + x\) with \(x + h\), leading to \(f(x + h) = -(x + h)^{2} + (x + h)\).

Step 2: Simplify \(f(x + h)\)

Expand the square and simplify to get \(f(x + h) = -x^{2} - 2hx - h^{2} + x + h\).

Step 3: Form the Difference Quotient

The difference quotient formula is \(\frac{f(x+h)-f(x)}{h}\), substitute \(f(x + h)\) and \(f(x)\), we get: \(\frac{-x^{2} - 2hx - h^{2} + x + h - (-x^{2} + x)}{h}\).

Step 4: Simplify the Difference Quotient

Simplify the numerator, leading to: \(\frac{-2xh - h^{2} + h}{h}\). This can be simplified further, by dividing each term in the numerator by \(h\), resulting in: \(-2x - h + 1\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial that can be written in the standard form \( f(x) = ax^2 + bx + c \), where \( a \), \( b \) and \( c \) are constants, and \( a \eq 0 \). These functions produce a parabola when graphed, with the direction of the opening depending on the sign of \( a \).

In the exercise provided, we are dealing with the quadratic function \( f(x) = -x^2 + x \), which is a downward opening parabola because the coefficient \( a \) (in this case \( -1 \) before \( x^2 \)) is negative. Quadratic functions often arise in problems involving area, projectile motion, and situations involving maximum or minimum values. Understanding how to handle and simplify expressions within quadratic functions is essential for solving many algebraic problems.

Function Simplification

Function simplification is the process of reducing an algebraic expression to its simplest form. This can involve combining like terms, factoring, expanding expressions, or canceling common factors.

For example, when finding the difference quotient for a quadratic function like \( f(x) = -x^2 + x \), the process involves function simplification to make the expression of \( f(x+h) \) more manageable to work with. In our exercise, after substituting \( x+h \) into the function and expanding, we simplify to get \( f(x + h) = -x^2 - 2hx - h^2 + x + h \).

Following the steps to find the difference quotient requires simplifying the expression further until we can cancel out the common factor, which in this case is \( h \), leading to a simplified result. Simplification makes it easier to understand the behavior of the function and to compute the final value of expressions, such as the slope of the tangent line to a curve at a point, represented by the difference quotient.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and arithmetic operations. They represent quantities in algebraic form and can be simple or complex, depending on the number of terms and the involved operations. In our exercise, the algebraic expression \( -x^2 - 2hx - h^2 + x + h \) represents the value of the quadratic function \( f(x) \) when \( x \) is increased by \( h \).

The process of dealing with algebraic expressions often entails applying the distributive property, combining like terms, and simplifying terms. When working with difference quotients, it is crucial to manipulate these expressions carefully – ensuring that all algebraic steps such as expansion, factorization, and simplification are carried out correctly. Proper manipulation of algebraic expressions is foundational to algebra and is applicable to higher-level mathematics and various real-world problems.