Problem 68 - College Algebra Essentials

Question

find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=-3 x^{2}+2 x-1 $$

Step-by-Step Solution

Answer

The simplified difference quotient for the given function is \(-6x - 3h + 2\).

Step 1: Plug into difference quotient

To start, substitute the equation \( f(x) = -3x^{2} + 2x -1 \) into the formula for the difference quotient. Make sure to replace every x in the f(x) function with \( x+h \). That would give the following: \n\(\frac{-3(x+h)^{2}+2(x+h)-1 - (-3x^{2}+2x-1)}{h}\).

Step 2: Simplify the numerator

After substitution, expand and simplify the terms in the numerator of the fraction: \n\(\frac{-3(x^2 + 2xh + h^2) + 2x + 2h - 1 - (-3x^{2}+2x-1)}{h}\).\n Simplifying this gives: \n\(\frac{-3x^2 - 6xh - 3h^2 + 2x + 2h - 1 + 3x^{2} - 2x + 1}{h}\).\n Combine like terms to get: \n\(\frac{- 6xh - 3h^2 + 2h}{h}\).

Step 3: Cancel out the common factor h

After simplifying, cancel out the common factor \(h\) from the numerator and the denominator of the fraction. That would result in: \n \(-6x - 3h + 2\).

Step 4: Provide the final simplified form of the difference quotient

Lastly, rewrite the expression in the standard form as the final answer which is: \n \(-6x - 3h + 2\).

Key Concepts

Algebraic Functions

Algebraic functions serve as fundamental building blocks in mathematics. In simple terms, these functions are formed using basic operations such as addition, subtraction, multiplication, division, and taking roots. Algebraic functions often involve variables that range over numbers, and they include polynomial functions, rational functions, and others.

When studying difference quotients in an algebraic context, you engage in understanding how functions change. The difference quotient itself depicts the average rate of change of a function over a small interval. This is essential in calculus as it forms the foundation for explaining derivatives, which represent instantaneous rates of change.

• In the problem given, you start with an algebraic function, specifically a polynomial, before substituting it into the difference quotient formula.
• The substitution process involves replacing the variable in the original function with an expression involving the increment variable \(h\).

Understanding algebraic functions in this way not only gives insight into their structure but also prepares you for more advanced concepts like calculus.

Simplification of Expressions

Simplifying expressions is a crucial skill in both algebra and calculus. It involves reducing expressions to their simplest form. This process makes calculations easier and more efficient.

When simplifying expressions in a complex formula like the difference quotient, there are several key steps:

• Expand all terms, making sure to apply distribution laws properly. This means carefully addressing each part of the expression where multiplication and addition are involved.
• Combine like terms meticulously. This step is crucial for reducing the equation to simpler components.
• Cancel out common factors where possible. In our exercise, the variable \(h\) in the numerator and denominator is canceled after simplification.

By mastering simplification, you enhance your algebra skills significantly, as it allows you to tackle more complex functions and better understand their behavior and transformations.

Polynomial Functions

Polynomial functions are among the simplest forms of algebraic functions, yet they are highly versatile and significant in mathematics. A polynomial function consists of terms, each including a coefficient and a variable raised to a non-negative integer exponent.

In the exercise, the function \(f(x) = -3x^2 + 2x - 1\) is a second-degree polynomial (quadratic function). Here are some important characteristics:

• Each term in a polynomial function represents a part of the whole picture of the function's graph.
• The degree of the polynomial provides insights into the number of roots and the behavior of the graph at infinity.
• Polynomials are smooth and continuous, which makes them crucial tools in calculus for approximating more complex functions.

Understanding polynomial functions is vital, not just for carrying out tasks such as calculating the difference quotient but also for predicting how changes in expressions affect graph shapes and behaviors. These skills are foundational for advanced studies in mathematics and applied sciences.