Problem 84 - A Graphical Approach to College Algebra

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) ) for each function \(f\). Simplify completely. $$f(x)=-2 x^{3}$$

Step-by-Step Solution

Answer

The difference quotient is \(-6x^2 - 6xh - 2h^2\).

Step 1: Determine f(x + h)

First, find the expression for \( f(x + h) \) by substituting \( x + h \) into the function \( f(x) = -2x^3 \). This gives us \( f(x + h) = -2(x + h)^3 \).

Step 2: Expand (x + h)^3

Expand the expression \((x + h)^3\) using the binomial theorem. Thus, \((x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\).

Step 3: Substitute and Expand f(x + h)

Substitute the expanded form of \((x + h)^3\) into \( f(x + h) = -2(x + h)^3 \). This gives us \( f(x + h) = -2(x^3 + 3x^2h + 3xh^2 + h^3) = -2x^3 - 6x^2h - 6xh^2 - 2h^3 \).

Step 4: Calculate f(x+h) - f(x)

Subtract \( f(x) = -2x^3 \) from \( f(x + h) = -2x^3 - 6x^2h - 6xh^2 - 2h^3 \). This results in \( f(x+h) - f(x) = (-2x^3 - 6x^2h - 6xh^2 - 2h^3) - (-2x^3) = -6x^2h - 6xh^2 - 2h^3 \).

Step 5: Form the Difference Quotient

Insert \( f(x+h) - f(x) \) into the difference quotient formula: \( \frac{f(x+h) - f(x)}{h} = \frac{-6x^2h - 6xh^2 - 2h^3}{h} \).

Step 6: Simplify the Difference Quotient

Divide each term in the numerator by \( h \). This simplifies the quotient to \( -6x^2 - 6xh - 2h^2 \).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that allows us to expand expressions of the form \((a+b)^n\) into a sum involving terms of the form \(a^j b^{n-j}\). This is particularly useful when dealing with polynomial expressions that need to be simplified or expanded. The theorem is expressed as follows:

• The expansion of \((a + b)^n\) consists of terms like \(\binom{n}{k} a^{n-k} b^k\).
• \(\binom{n}{k}\) represents the binomial coefficient, which can be calculated as \(\frac{n!}{k!(n-k)!}\).
• This is used to expand and simplify polynomial expressions efficiently.

In our exercise, we needed to expand \((x+h)^3\), which is a direct application of the theorem:

• Using the binomial theorem results in \(x^3 + 3x^2h + 3xh^2 + h^3\).
• This form makes it easier to substitute back into the larger expression for further simplification steps.

Understanding this helps simplify complex algebraic expressions by breaking them down into more manageable terms.

Polynomial Functions

Polynomial functions are expressions that involve variables raised to whole number powers and include coefficients. They can take many forms, such as linear, quadratic, cubic, and so on, based on the highest power of the variable:

• A linear polynomial has a highest degree of 1, such as \(f(x)=3x+2\).
• A quadratic polynomial has a degree of 2, like \(f(x)=x^2-4x+4\).
• Cubic polynomials, such as \(f(x)=-2x^3\), have a degree of 3.

In our exercise, the polynomial function \(f(x)=-2x^3\) is cubic and involves a third degree term only. Such polynomials are important as they can represent various physical systems and real-world scenarios. In working through problems like the one given, understanding how to perform operations on polynomial functions, including substitution and manipulation, is crucial.

• Polynomials can often be simplified using algebraic techniques like factorization and expansion.
• They are foundational in calculus for studying limits, derivatives, and integrals.

Algebraic Simplification

Algebraic simplification involves reducing expressions to their simplest form, making them easier to work with. This process often involves combining like terms, factoring, and reducing fractions. It is a vital skill in solving algebraic equations and working with polynomial functions:

• Combine like terms to reduce the number of terms in an expression.
• Applying distributive properties to simplify expressions with parentheses.
• Simplifying fractions by dividing out common factors.

For the difference quotient in this exercise, we used simplification to reduce the expression \(\frac{-6x^2h - 6xh^2 - 2h^3}{h}\). Each term was divided by \(h\), leading to a simpler expression: \(-6x^2 - 6xh - 2h^2\). This step is crucial for analyzing limits and understanding behavior of functions as the variables change.

• Simplification ensures that the expressions are interpreted easily and efficiently, especially in calculus for derivative calculations.
• It helps prevent errors that might occur from handling unnecessarily complicated expressions.

In summary, mastering algebraic simplification is key to efficiently handling mathematical expressions and problems.