Problem 67 - College Algebra Essentials

Question

find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=-2 x^{2}+5 x+7 $$

Step-by-Step Solution

Answer

The simplified difference quotient for the function \(f(x)=-2x^{2}+5x+7\) is \(-4x -2h +5\).

Step 1: Substitute function into the difference quotient

Insert the function \(f(x) = -2x^{2} + 5x + 7\) into the difference quotient formula. This results in: \[\frac{f(x+h)-f(x)}{h} = \frac{(-2(x+h)^{2}+5(x+h)+7)-(-2x^{2}+5x+7)}{h}\]

Step 2: Simplify the numerator

Expand and simplify the expression in the numerator (top part) of the fraction. This results in: \[\frac{-2(x^{2}+2xh+h^{2})+5x+5h+7-(-2x^{2}+5x+7)}{h} = -4xh -2h^{2} +5h\].

Step 3: Divide by h

Then, divide each term in the numerator by \(h\) (since \(h \neq 0\)). This results in: \[-4x -2h +5\].

Step 4: Simplify Further

Simplify the obtained expression as much as possible. In this case, no further simplification is possible, so this is the final difference quotient.

Key Concepts

Understanding Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to different powers. These powers are whole numbers, and each term in the polynomial is a product of a constant and a power of the variable. The general form of a polynomial is given as:

• \[a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\]

Here, each "\(a\)" stands for a coefficient, while "\(x\)" is the variable raised to various powers.
For the function provided in the exercise, the polynomial is \(-2x^2 + 5x + 7\). This is a second-degree polynomial, as the highest power of \(x\) is squared.
Polynomial functions are widely used to describe various phenomena in science and engineering due to their smooth and continuous nature.
Their behavior is influenced by the coefficients and the degree of the polynomial.

Demystifying Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations (like addition, subtraction, multiplication, and division). They form the building blocks of algebra.
In our exercise, we work with algebraic expressions by inserting them into formulas like the difference quotient.

• The formula involves combining expressions and simplifying them, which often requires expanding them first.
• Essential algebraic skills include understanding how to factor expressions, use the distributive property, and combine like terms.

In the step-by-step solution, we used these skills to expand \((-2(x+h)^2 + 5(x+h) + 7)\). We then simplified the terms by distributing and combining like terms, leading to an expression \(-4xh - 2h^2 + 5h\) in the numerator.
Breaking down such expressions step by step is crucial to understanding and simplifying complex algebraic problems.

Exploring Calculus Concepts

Calculus introduces the concept of limits, derivatives, and integrals, offering powerful tools for analyzing changing quantities.
The difference quotient is a central component in calculus, acting as a precursor to the derivative.

• The formula \(\frac{f(x+h)-f(x)}{h}\) measures the average rate of change of a function \(f(x)\) over an interval \(h\).
• As \(h\) approaches zero, this average rate of change can be used to find the instantaneous rate of change, or the derivative of the function.

In the given exercise, the difference quotient is simplified to \(-4x - 2h + 5\).
Understanding this concept is essential because derivatives are fundamental in solving numerous real-world applications like velocity, slopes of curves, and optimizing functions.
For instance, when analyzing motion, the derivative provides the velocity of an object at any point in time.