Problem 66 - Algebra and Trigonometry

Question

\(65-70\) m Simplify the fractional expression. (Expressions like these arise in calculus.) $$ \frac{(x+h)^{-3}-x^{-3}}{h} $$

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-3x^2 - 3xh - h^2}{(x+h)^3x^3}\).

Step 1: Understand the Expression

The given expression is \( \frac{(x+h)^{-3} - x^{-3}}{h} \). This is a difference quotient, commonly seen in calculus, representing the difference of two expressions raised to a negative power, divided by \( h \).

Step 2: Apply Powers of Binomials

The term \((x+h)^{-3}\) can be expanded using the binomial theorem or treated conceptually to understand that it's equivalent to \( \frac{1}{(x+h)^3} \).

Step 3: Simplify the Numerator

The numerator \((x+h)^{-3} - x^{-3}\) can be rewritten as \(\frac{1}{(x+h)^3} - \frac{1}{x^3}\). To subtract these fractions, find a common denominator, which is \((x+h)^3x^3\).

Step 4: Combine the Fractions

Rewrite the numerator with a common denominator: \(\frac{x^3 - (x+h)^3}{(x+h)^3x^3}\). Use the formula for the difference of cubes: \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\).

Step 5: Substitute the Difference of Cubes

Now, substitute \(a = x\) and \(b = (x+h)\) into the formula, yielding \(x^3 - (x+h)^3 = (-h)(2x^2 + 2xh + h^2)\). Substitute back into the fraction.

Step 6: Factor Out \(h\)

We now have \(\frac{h(-3x^2 - 3xh - h^2)}{(x+h)^3x^3}\). You can factor \(h\) out of the numerator and cancel it with the \(h\) in the denominator.

Step 7: Simplified Expression

After factoring \(h\) and canceling, we obtain \(\frac{-3x^2 - 3xh - h^2}{(x+h)^3x^3}\) as the simplified expression.

Key Concepts

Difference Quotient

In calculus, the difference quotient is a crucial concept used to define the derivative of a function. It provides a way to calculate the slope of the secant line between two points on a curve. This becomes particularly handy when working towards finding the instantaneous rate of change of a function at a single point. The general form of the difference quotient for a function \(f(x)\) is:

• \(\frac{f(x+h) - f(x)}{h}\)

In the context of the given expression, \(\frac{(x+h)^{-3} - x^{-3}}{h}\), we are essentially evaluating the change in the function \((x+h)^{-3}\) as \(h\) approaches zero, giving us insights into the behavior of the function at \(x\). To simplify, you have to carefully perform algebraic manipulations while keeping this form in mind.

Power of Binomials

The term \((x+h)^{-3}\) represents the power of a binomial raised to a negative exponent. In algebra, this involves considering both positive and negative powers, which can be understood better through the Binomial Theorem for expansion. However, in this case, it simplifies to:

• \((x+h)^{-3} = \frac{1}{(x+h)^3}\)

This transformation indicates the inverse cubic power of the sum \((x+h)\). Recognizing this relationship is key when simplifying expressions like the one in our exercise. It ensures that the subsequent algebraic steps, like finding a common denominator, are completed efficiently and accurately.

Difference of Cubes

The formula for the difference of cubes is an essential algebraic tool. It helps to simplify expressions like \(a^3 - b^3\). The formula is:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

In our exercise, identifying that \(a = x\) and \(b = x+h\) allows us to expand \(x^3 - (x+h)^3\) effectively. The difference of cubes creates factors that can subsequently be simplified, making it significantly simpler to deal with operations like finding common denominators or cancelling terms. By substituting these expressions into the fraction, you enable further simplification.

Calculus

Calculus revolves heavily around the concepts of derivatives and integrals, focusing on rates of change and accumulation, respectively. The calculation of the derivative often starts with concepts like the difference quotient, enabling the step-by-step approach to determine the slope of the tangent at a point. This approach is rooted in limits, where expressions of the form \(\frac{(x+h)^{-3} - x^{-3}}{h}\) become more significant as \(h\) approaches zero.Simplification practices, including the use of binomial expansions and cube differences, ensure the process remains accurate and compact. Comprehending these expressions' behavior is central to solving calculus problems, as it forms the basis of understanding change over infinitesimal distances.