Problem 50 - Calculus and its applications

Question

Find the simplified difference quotient for each function listed. $$ f(x)=x^{5} $$

Step-by-Step Solution

Answer

The simplified difference quotient is \( 5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4 \).

Step 1: Understanding the Difference Quotient

The difference quotient is defined as \( \frac{f(x+h) - f(x)}{h} \). It is used to find the derivative of a function in its limit form as \( h \to 0 \). For this exercise, we will simplify this expression for the given function \( f(x) = x^5 \) without taking the limit.

Step 2: Substituting into the Formula

First, substitute \( f(x) = x^5 \) into the difference quotient formula. This gives us: \( \frac{(x+h)^5 - x^5}{h} \). Our task now is to expand \( (x+h)^5 \) and simplify.

Step 3: Expanding \((x+h)^5\)

The expansion of \( (x+h)^5 \) can be done using the Binomial Theorem: \( (x+h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5 \). Substitute this expanded form back into the difference quotient.

Step 4: Substituting the Expansion

After substituting the expansion, the expression becomes: \( \frac{x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5 - x^5}{h} \). Notice that the \( x^5 \) terms cancel out.

Step 5: Simplifying the Expression

The expression simplifies to \( \frac{5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5}{h} \). Factor \( h \) out from the numerator to get \( h(5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4) \).

Step 6: Canceling \( h \) and Final Simplification

Cancel out \( h \) in the numerator and the denominator, yielding \( 5x^4 + 10x^3h + 10x^2h^2 + 5xh^3 + h^4 \). This is the simplified difference quotient for \( f(x) = x^5 \).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in mathematics that allows us to expand expressions of the form \((x + h)^n\) easily. It's especially helpful when dealing with power expressions which need expanding, such as \((x+h)^5\). The general binomial expansion formula is given by \[ (x+h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k \] Here, \(\binom{n}{k}\) are the binomial coefficients, calculated as \(\frac{n!}{k!(n-k)!}\). This approach significantly simplifies computations and makes it more manageable to see how terms distribute in expressions involving sums of powers. Applying this theorem to our exercise:

• We have \((x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5\).
• Each term represents a combination where powers of \(x\) decrease and powers of \(h\) increase.
• This method saves time in algebraic expansions, especially when simplifying expressions like the difference quotient.

Calculus

Calculus is the mathematical study that focuses on change, using principles such as derivative to understand how functions behave. The difference quotient is an introductory concept in calculus, providing a gateway to the derivative. Understanding the definition, it forms the expression: \[ \frac{f(x+h) - f(x)}{h} \]. In calculus, this expression is crucial to finding the slope of the tangent line at a particular point on the function, thus illustrating instantaneous change. Here's the connection in the exercise:

• We compute \( f(x+h) \) and substitute it in the difference quotient formula.
• After expansion and simplification of the polynomial, we can easily cancel terms and factor out \(h\) effectively.
• This often leads us to understand how calculus helps in determining the behavior and rates of change of functions.

Ultimately, calculus enables us to simplify complex expressions and glean significant insights about functions' geometrical and dynamic properties.

Function Simplification

Function simplification is a fundamental mathematical practice where complex expressions and formulas are reduced to their simpler forms without changing their values. In this exercise, simplification occurs at a couple of critical points:

• First, we use the binomial theorem to expand \((x+h)^5\), a necessary step for substituting into the difference quotient.
• Then after substituting, further simplification occurs by eliminating common terms, such as cancelling out \(x^5\) from the expression.
• Finally, factoring out \(h\) from the polynomial in the numerator allows for a crucial step: cancelling \(h\) from the fraction which is pivotal for reaching the simplified form.

This process of simplification is vital in making algebraic manipulations manageable and attainable. It reduces errors and highlights core relationships in mathematical expressions. By practicing simplification, students can solve problems more efficiently and understand the underlying structures of functions.