Problem 70 - Precalculus with Limits

Question

In Exercises 67 - 72, expand the expression in the difference quotient and simplify. \( \dfrac{f\left(x + h\right) - f\left(x\right)}{h} \quad \quad \) Difference quotient \( f(x) = x^8 \)

Step-by-Step Solution

Answer

After expanding and simplifying the difference quotient with \( f(x) = x^8 \), the result is \( 8x^7 + 28x^6h + 56x^5h^2 + 70x^4h^3 + 56x^3h^4 + 28x^2h^5 + 8xh^6 + h^7 \)

Step 1: Expand the expression

Start by substituting \( f(x) = x^8 \) into the difference quotient. It will become \( \dfrac{(x+h)^8 - x^8}{h} \). Now expand \( (x+h)^8 \) using the binomial theorem.

Step 2: Apply the binomial theorem

The binomial theorem allows us to expand \( (x + h)^8 \) into \( x^8 + 8x^7h + 28x^6h^2 + 56x^5h^3 + 70x^4h^4 + 56x^3h^5 + 28x^2h^6 + 8xh^7 + h^8 \). Substituting back into the difference quotient we get \( \dfrac{x^8 + 8x^7h + 28x^6h^2 + 56x^5h^3 + 70x^4h^4 + 56x^3h^5 + 28x^2h^6 + 8xh^7 + h^8 - x^8}{h} \)

Step 3: Simplify the expression

Notice that \( x^8 \) in the numerator cancels out. Divide each term in the numerator by \( h \) leaving us with the answer: \( 8x^7 + 28x^6h + 56x^5h^2 + 70x^4h^3 + 56x^3h^4 + 28x^2h^5 + 8xh^6 + h^7 \)

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool in mathematics that allows us to expand expressions of the form \((x + a)^n\) into a series of terms. The theorem is incredibly useful, particularly in calculus, for handling polynomials raised to powers where direct multiplication would be cumbersome. It states that:

• \((x + a)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}a^k \)

Here, \(\binom{n}{k}\) represents binomial coefficients, which can be calculated as \(\frac{n!}{k!(n-k)!}\).
In our exercise, we used the binomial theorem to expand \((x + h)^8\). This resulted in a series of terms, each involving powers of \(x\) and \(h\). By understanding this expansion, it becomes easier to perform algebraic manipulations and simplifications, especially in calculus where such forms are common.

Calculus

Calculus comes into play primarily through the concept of a limit, often applied to the difference quotient \( \frac{f(x+h) - f(x)}{h} \). This difference quotient is fundamental as it represents the average rate of change of the function \(f(x)\) over a small interval \(h\). As \(h\) approaches 0, this expression determines the instantaneous rate of change or the derivative, denoted as \(f'(x)\).

• The process we explored—expanding, simplifying, and then dividing by \(h\)—is one-many students encounter repeatedly.
• These steps help conclude with a polynomial representing a derivative of the function \(f(x)\).

In practice, after simplification, this derivative provides insights into the function’s behavior, such as the slope of its tangent at any point \(x\). The calculation of derivates is a cornerstone of differential calculus.

Simplification of Expressions

Simplification of expressions is a fundamental skill in both algebra and calculus. After expanding \((x+h)^8\) using the binomial theorem, we simplified the resulting expression to obtain a manageable form for analysis and interpretation.

• First, recognize and eliminate any like terms within the mathematical expression.
• In our case, \(x^8\) terms canceled out during subtraction in the numerator.

Next, dividing each term by \(h\) at the end isolates the expression in a simpler form: \(8x^7 + 28x^6h + 56x^5h^2 + \ldots + h^7\).
This simplification is crucial not only for making calculations easier but also for enhancing our understanding of the algebraic structure and behavior of mathematical expressions, especially when used to derive and understand limits and derivatives in calculus.