Problem 67 - 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 \( f(x) = x^3 \)

Step-by-Step Solution

Answer

The expanded and simplified difference quotient for \( f(x) = x^3 \) is \( 3x^2 \).

Step 1: Find f(x+h)

The first step is to substitute \( x + h \) into the function \( f(x) = x^3 \). This yields: \( f(x+h) = (x+h)^3 \).

Step 2: Expand the expression

Next, expand \( (x + h)^3 \) using the binomial theorem, which results in \( x^3 + 3x^2h + 3xh^2 + h^3 \).

Step 3: Substitute into the Difference Quotient

Now, substitute \( f(x+h) \) and \( f(x) \) into the difference quotient expression. This gives: \( \dfrac{(x^3 + 3x^2h + 3xh^2 + h^3) - x^3}{h} \).

Step 4: Simplify the expression

The exponent part of the difference quotient now simplifies to \( 3x^2 + 3xh + h^2 \).

Step 5: Evaluate the limit as h approaches zero

Finally, calculate the limit of the expression as \( h \) approaches zero. As \( h \) approaches zero, all terms containing \( h \) in the equation will also approach zero. Therefore, the final simplified difference quotient yields \( 3x^2 \).

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful method for expanding expressions that are raised to a power. It provides a formula for expanding any power of a binomial, such as \((x + h)^3\). This expansion is crucial in calculus, particularly when working with derivatives and difference quotients.

The Binomial Theorem states that:\[(x + h)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} h^k\]where \( \binom{n}{k} \) is a binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

In practice, this means that every term in the expansion will include a combination of powers from both \(x\) and \(h\). For our task, expanding \((x+h)^3\) results in:

• The first term, \(x^3\), from \(x^3 \times h^0\),
• The second term, \(3x^2h\), from \(\binom{3}{1} x^2h\),
• The third term, \(3xh^2\), from \(\binom{3}{2} xh^2\),
• The fourth term, \(h^3\), from \(h^3\).

This expansion becomes essential when dealing with difference quotients, as it sets the stage for simplifying expressions before limits are calculated.

Limit Calculations

Limit calculations are central to calculus, especially when finding derivatives using difference quotients. A limit explores what happens to a function as the input approaches a certain value. In our example, this means examining how the difference quotient changes as \(h\) approaches zero.

After simplifying the difference quotient to \((3x^2 + 3xh + h^2)\), the goal is to evaluate the limit:\[\lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\]Understanding limits is crucial since they help determine the instantaneous rate of change of a function, which is the derivative.

In the example, since all terms contain \(h\), we notice that as \(h\) trends towards zero, these terms vanish, simplifying our expression effectively. It leaves us with the derivative \(3x^2\), a result born from the binomial expansion and limit evaluation.

Polynomial Functions

Polynomial functions, like \(f(x) = x^3\), are mathematical expressions that include variables raised to whole-number powers. They're a fundamental concept in algebra and calculus due to their straightforward behavior and properties.

Each polynomial function comprises:

• Terms, which are the distinct components, e.g., \(x^3\),
• Coefficients, like 1 in \(x^3\), multiplied by the variables,
• Degrees, determined by the highest exponent, such as 3 in \(x^3\).

When engaged with the difference quotient, these functions allow for simplification, often enabling more manageable limit calculations.

Moreover, polynomial functions maintain smooth graphs and continuous derivatives, making them ideal for calculus studies. Given their properties, they're frequently used in modeling and solving real-world problems, providing a solid foundation for understanding more complex mathematical operations.