Problem 82 - College Algebra with Modeling and Visualization

Question

Complete the following for the given \(f(x)\) (a) Find \(f(x+h)\) (b) Find the difference quotient of \(f\) and simplify. $$ f(x)=1-x^{3} $$

Step-by-Step Solution

Answer

(a) \( f(x+h) = 1 - x^3 - 3x^2h - 3xh^2 - h^3 \), (b) \( -3x^2 - 3xh - h^2 \)

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

To find \(f(x+h)\), substitute \(x+h\) into the function \(f(x) = 1 - x^3\).\[ f(x+h) = 1 - (x+h)^3 \]

Step 2: Expand the Binomial (x + h)^3

Use the binomial theorem to expand \((x+h)^3\):\[(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\]Substitute back into the function:\[ f(x+h) = 1 - (x^3 + 3x^2h + 3xh^2 + h^3) \]

Step 3: Simplify Expression for f(x + h)

Simplify the expression by distributing the negative sign:\[ f(x+h) = 1 - x^3 - 3x^2h - 3xh^2 - h^3 \]

Step 4: Set Up the Difference Quotient

The difference quotient for \(f(x)\) is given by:\[ \frac{f(x+h) - f(x)}{h} \]

Step 5: Substitute and Simplify Difference Quotient

Substitute \(f(x+h)\) and \(f(x) = 1 - x^3\) into the difference quotient:\[ \frac{1 - x^3 - 3x^2h - 3xh^2 - h^3 - (1 - x^3)}{h} \]Simplify:\[ \frac{-3x^2h - 3xh^2 - h^3}{h} \]

Step 6: Simplify by Canceling h

Factor out \(h\) from the numerator and cancel it with the denominator:\[ \frac{h(-3x^2 - 3xh - h^2)}{h} = -3x^2 - 3xh - h^2 \]

Step 7: Find the Simplified Difference Quotient

The simplified difference quotient is:\[ -3x^2 - 3xh - h^2 \]

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions that involve the sum of powers of a variable, each multiplied by a coefficient. In simple terms, they are functions that consist of terms involving variables raised to whole number exponents and their respective coefficients. Here are some key points to remember about polynomial functions:

• The general form of a polynomial function is: \( f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \).
• The highest exponent of the variable determines the degree of the polynomial.
• Each term consists of a coefficient and a variable raised to an exponent, like \(a_nx^n\).

For instance, in the exercise, the function \(f(x) = 1-x^3\) is a simple polynomial function of degree 3. It consists of two terms, a constant (1) and a cubic term (-\(x^3\)). Understanding the structure of polynomial functions helps in solving various mathematical problems and expanding expressions, like computing \(f(x+h)\).

Binomial Expansion

Binomial expansion is an essential part of working with polynomial functions, especially when dealing with expressions of the form \((x+h)^n\). The binomial theorem gives us a way to expand these expressions efficiently:

• It states that \((x + y)^n\) can be expanded into a series of terms in the form of \(a \cdot b\), where each term is a combination of the original variables raised to different powers.
• For example, \((x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3\).
• The coefficients of the expanded terms correspond to the binomial coefficients, found in Pascal's Triangle or by the formula for combinations \(\binom{n}{k}\).

In our exercise, expanding \((x+h)^3\) is crucial because it allows for further simplification of \(f(x+h)\). Knowing how to perform a binomial expansion helps simplify functions and is particularly useful in calculus when finding limits and derivatives.

Function Simplification

Simplifying functions is a crucial step in calculus and algebra that involves rewriting expressions in a simpler or reduced form without changing their value. This process is important as it makes equations easier to handle and interpret.Here are the steps typically involved in function simplification:

• Substitution: Replace variables or constants as instructed, like finding \(f(x+h)\) by substituting \(x+h\) into \(f(x)\).
• Expansion: Use techniques such as the binomial theorem to expand complex expressions into their expanded forms.
• Distribution: Apply distribution laws to remove parentheses and combine like terms, simplifying the function further.
• Reduction: Cancel common terms or factors across the numerator and denominator, reducing the expression to its simplest form.

In the given solution, the simplification step for the difference quotient of \(f(x)\) involves subtracting \(f(x)\) from \(f(x+h)\), and then canceling the \(h\) factor from both the numerator and denominator. This leaves a simpler expression that helps in evaluating limits or finding derivatives, which are key concepts in calculus.