91Ó°ÊÓ

Simplifying expressions is a crucial skill in algebra and calculus that helps in understanding complex functions and their behavior. In the context of the difference quotient, simplifying the expression is all about breaking down and eliminating unnecessary complexity.

The original expression for the difference quotient in this exercise is:\[ \frac{f(x+h) - f(x)}{h} \]For the function \( f(x) = 1 - x^3 \), you find \( f(x+h) = 1 - (x+h)^3 \). With this substitution, your task is to simplify: \[ \frac{(1 - (x+h)^3) - (1 - x^3)}{h} \]The simplification process involves:
- Expanding \((x+h)^3\) to get terms like \(x^3 + 3x^2h + 3xh^2 + h^3\).
- Canceling out terms such as \(1 - 1\) and \(x^3 - x^3\).
This results in a simplified numerator of\(-3x^2h - 3xh^2 - h^3\).

The last step in this simplification process is to factor out \(h\). The expression becomes: \[ h(-3x^2 - 3xh - h^2) \]. Finally, dividing by \(h\) cancels it out, leaving: \[ -3x^2 - 3xh - h^2 \].
Through these steps, what started as a complex equation is reduced to a simpler form that offers more straightforward mathematical processing.

The concept of derivatives in calculus is closely related to the difference quotient. A derivative represents the rate of change of a function and can be understood as the "instantaneous slope" of the function at any given point.

In this exercise, the difference quotient provides a foundational step toward finding the derivative, which is particularly useful when dealing with polynomial functions like \( f(x) = 1 - x^3 \). By simplifying the difference quotient \[ \frac{f(x+h) - f(x)}{h} \] as \(-3x^2 - 3xh - h^2\), you're approaching the derivative of the function.

Consider any point \(x\) and let the increment \(h\) approach zero. As \(h\) gets smaller, closer to zero, the expression \(-3x^2 - 3xh - h^2\) tends toward the derivative of the function at that point.

So the simplified expression gives you an approximate slope, and as \(h\) approaches zero, you get the exact slope or derivative which is \(-3x^2\). This understanding helps in various practical calculations and in predicting how a function behaves.

Numerical calculations make the application of mathematical expressions more tangible. In this exercise, we need to calculate the simplified form of the difference quotient for different values of \(h\), specifically \(h = 2\), \(1\), \(0.1\), and \(0.01\) with \(x = 5\).

Using the simplified expression \(-3x^2 - 3xh - h^2\), let's compute these values:

• For \(x = 5, h = 2\):
  Expression becomes \(-3(5)^2 - 3(5)(2) - (2)^2 = -75 - 30 - 4 = -109\).
• For \(x = 5, h = 1\):
  Expression becomes \(-3(5)^2 - 3(5)(1) - (1)^2 = -75 - 15 - 1 = -91\).
• For \(x = 5, h = 0.1\):
  Expression becomes \(-3(5)^2 - 3(5)(0.1) - (0.1)^2 = -75 - 1.5 - 0.01 = -76.51\).
• For \(x = 5, h = 0.01\):
  Expression becomes \(-3(5)^2 - 3(5)(0.01) - (0.01)^2 = -75 - 0.15 - 0.0001 = -75.1501\).

Observing these calculations, it's clear that as \(h\) gets smaller, the result approaches \(-75\), which resonates with the concept of derivative. This emphasizes the importance of numerical calculations in verifying the behavior of functions as parameters change.