Problem 56 - Single Variable Calculus

Question

Let \(f(x)=x^{3}-x\). a. Find \(f^{\prime}(x)\). b. Plot the graphs of \(f^{\prime}\) and \(g\), where $$g(x)=\frac{\left[(x+0.01)^{3}-(x+0.01)\right]-\left(x^{3}-x\right)}{0.01}$$ using a common viewing window. Is the result expected? Explain.

Step-by-Step Solution

Answer

a. The derivative of \(f(x) = x^3 - x\) is \(f^{\prime}(x) = 3x^2 - 1\). b. After plotting the graphs of \(f^{\prime}(x)\) and \(g(x)\), we observe that they are very similar, almost coinciding. This result is expected since \(g(x)\) is a numerical approximation of the derivative of \(f(x)\), illustrating the rate of change of \(f(x)\) using numerical methods. Meanwhile, \(f^{\prime}(x)\) gives the exact value of the derivative using calculus.

Step 1: Find the derivative of \(f(x)\)

To find the derivative \(f^{\prime}(x)\), we apply the power rule. The power rule states that \(\frac{d}{dx}(x^n)=nx^{n-1}\) for any power n: \[f^{\prime}(x) = \frac{d}{dx}(x^3 - x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(x)\] Applying the power rule for both terms: \[f^{\prime}(x) = 3x^2 - 1\]

Step 2: Plot the graphs of \(f^{\prime}(x)\) and \(g(x)\)

Using a graphing calculator or software like Desmos, we are going to plot the graphs of both functions with a common viewing window. The functions to be plotted are: 1. The derivative of \(f(x): f^{\prime}(x) = 3x^{2} - 1\) 2. The function \(g(x):\) \[g(x) = \frac{\left[(x+0.01)^{3}-(x+0.01)\right]-\left(x^{3}-x\right)}{0.01}\] Once the plotting is done, we observe and compare the graphs of these two functions.

Step 3: Analyze the Graphs

When we plot the graphs of both functions \(f^{\prime}(x)\) and \(g(x)\), we can see that they are similar, almost coinciding. This is because \(g(x)\) is a numerical approximation of the derivative of \(f(x)\), which is given by the difference quotient formula. The difference quotient formula is commonly used as a finite difference approximation of derivatives. Since the graphs look very similar, the result is expected as both functions give us information about the rate of change of the function \(f(x)\). The function \(g(x)\) illustrates the concept of the derivative using numerical methods, while the function \(f^{\prime}(x)\) depicts the exact value of the derivative we found using calculus.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It essentially measures the rate at which a function's output value is changing at any given point. For a function \( f(x) \), its derivative, denoted as \( f'(x) \), is defined as a limit: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]This limit gives us the instantaneous rate of change of the function. The concept of derivatives is foundational in calculus as it extends to understanding motion, growth, and other applications where change is involved. In our exercise, the derivative \( f'(x) = 3x^2 - 1 \) illustrates how the given function \( f(x) = x^3 - x \) behaves or changes as \( x \) varies.

Power Rule

The power rule is a fundamental shortcut in calculus for finding the derivative of functions of the form \( x^n \). It states:

• For any real number \( n \), the derivative \( \frac{d}{dx}(x^n) = nx^{n-1} \).

This rule is quick and efficient, especially when working with polynomial functions. For example, to find the derivative of \( x^3 \), apply the power rule: multiply by the exponent (3), then decrease the exponent by one, resulting in \( 3x^2 \). Similarly, for \( x \) (which is \( x^1 \)), the derivative is \( 1 \cdot x^{0} = 1 \). In the exercise step, the power rule was applied to solve \( f'(x) = \frac{d}{dx}(x^3 - x) = 3x^2 - 1 \). This step is essential for simplifying and solving problems involving derivatives.

Finite Difference Approximation

Finite difference approximation is a method to estimate the derivative of a function using its values at specific points. This technique is particularly useful in numerical methods when an algebraic expression for the derivative is hard to find. The core idea is to use the difference quotient as:\[g(x) = \frac{f(x+h) - f(x)}{h}\]If \( h \) is small, \( g(x) \) approximates the derivative \( f'(x) \). In the exercise, \( g(x) \) was defined with \( h = 0.01 \):\[g(x) = \frac{(x + 0.01)^3 - (x + 0.01) - (x^3 - x)}{0.01}\]This formula computes differences between slightly shifted points to approximate \( f'(x) \). Since the step size \( h \) is small, the approximation is close to the exact derivative, helping illustrate numerical derivative approaches.

Numerical Methods

Numerical methods in calculus involve techniques for solving mathematical problems using approximations. These methods are especially helpful when exact solutions are difficult or impossible to derive analytically. They are often implemented using computers to perform the tedious calculations needed for precise approximations.

• Finite difference methods are a cornerstone of numerical differentiation, which is observed in our exercise when calculating \( g(x) \).
• Another common technique is the use of trapezoidal and Simpson's rules for integrals.
• Numerical methods are used extensively to solve real-world problems in engineering, sciences, and economics where analytic solutions are complex.

In our exercise, numerical methods, such as the finite difference approximation of \( g(x) \), serve to approximate the derivative of \( f(x) \). These methods are potent tools for analyzing changes and behaviors of functions when precise calculus approaches aren't readily accessible.