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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 \( f(x) = x^6 \)

Short Answer

Expert verified
The expansion and simplification of the difference quotient \( \dfrac{f\left(x + h\right) - f\left(x\right)}{h} \) for the function \( f(x) = x^6 \) is \( 6x^5 + 15x^4h + 20x^3h^2 + 15x^2h^3 + 6xh^4 + h^5 \).

Step by step solution

01

Insert the function into the difference quotient

To start off, substitute the function \( f(x) = x^6 \) into the difference quotient \( \dfrac{f\left(x + h\right) - f\left(x\right)}{h} \). The difference quotient becomes \( \dfrac{(x+h)^6 - x^6}{h} \).
02

Expand the terms

Expand the numerator of the quotient by using the binomial theorem, which says that \( (x + h)^6 = x^6 + 6x^5h + 15x^4h^2 + 20x^3h^3 + 15x^2h^4 + 6xh^5 + h^6 \). After expansion, the difference quotient becomes \( \dfrac{x^6 + 6x^5h + 15x^4h^2 + 20x^3h^3 + 15x^2h^4 + 6xh^5 + h^6 - x^6}{h} \).
03

Simplify the difference quotient

The terms \( + x^6 - x^6 \) in the numerator will cancel each other out. Additionally, each term in the numerator has a factor of 'h', which can be factored out. After factoring out 'h' from the numerator, we have \( h(6x^5 + 15x^4h + 20x^3h^2 + 15x^2h^3 + 6xh^4 + h^5) \). The 'h' in the numerator and the 'h' in the denominator will cancel out, leaving the simplification of the difference quotient as: \( 6x^5 + 15x^4h + 20x^3h^2 + 15x^2h^3 + 6xh^4 + h^5 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Simplifying Expressions
Understanding how to simplify expressions is a pivotal skill in algebra. It often involves combining like terms, factoring, and canceling. Simplifying can make complex equations more manageable and easier to work with.

For instance, in the difference quotient \(\dfrac{f(x + h) - f(x)}{h}\), simplification is key to finding the limit as \(h\) approaches zero. Initially, the expression may seem complicated, but by expanding and then canceling out identical terms on the numerator and denominator, we reveal the simplified expression that can be evaluated further.

Remember, the ultimate goal of simplification is to rewrite expressions into the simplest form possible. This often means isolating the unknown variable or getting to a point where no further reductions can be made. It is similar to cleaning up after cooking; you're left with only what is essential.
Binomial Theorem
The binomial theorem is a key player in expanding polynomial expressions. When we have an expression like \( (x + h)^6 \), it may seem daunting to expand manually. This is where the binomial theorem comes to the rescue, providing a systematic method for expansion.

The theorem expresses the result of raising a binomial to any power as a sum of terms in the form of \(C(n, k)x^{n-k}h^k\), where \(C(n, k)\) represents the binomial coefficients, or 'choose' function, giving us the count of ways to choose \(k\) elements from a set of \(n\) elements without considering the order.

So, for \( (x + h)^6 \), we use the theorem to quickly obtain the expanded form, including terms \( 6x^5h \), \( 15x^4h^2 \), and so on. It's a powerful toolkit for tackling polynomial expansion in difference quotients and beyond.
Polynomial Functions
Polynomial functions are expressions involving sums of powers of variables with constant coefficients. They are everywhere in mathematics and applied sciences. The expression \(x^6\) is a simple example of a polynomial function, where \(x\) is the variable, and \(6\) is the exponent showing the degree of the polynomial.

In our exercise, the difference quotient includes applying the polynomial function \(f(x + h)\) and \(f(x)\), indicating how the polynomial behaves under small changes in its input (the \(h\)). This is a foundational concept in calculus, specifically in defining the derivative of a function.

A good grasp of polynomial functions enables a deeper understanding of curves and can aid in solving a plethora of problems ranging from simple algebra to complex calculus. When we cancel out terms in the difference quotient, what remains is a simpler polynomial function that describes the rate at which \( f(x) \) changes with respect to \(x\). Polynomials are like the building blocks of algebraic expressions, and understanding their nature is crucial for advanced mathematical problem-solving.

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Most popular questions from this chapter

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