Problem 35 - Calculus with Concepts in Calculus

Question

Plot the graph of \(f^{\prime}\), and then use the Newton-Raphson method to approximate all values of \(c\) for which \((c, f(c))\) is an inflection point. Continue until the output of the calculator does not change. $$ f(x)=e^{x}-\sin x \text { for }-e \leq x<\infty $$

Step-by-Step Solution

Answer

Use Newton-Raphson on \(f''(x) = 0\) to find inflection at \(-2.073\).

Step 1: Find the first derivative

To start with, find the derivative of the function \(f(x) = e^x - \sin x\). Using the differentiation rules, we have:\[f'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(\sin x) = e^x - \cos x\]

Step 2: Find the second derivative

Next, find the second derivative of the function \(f(x)\) to locate the potential inflection points. Differentiate \(f'(x)\):\[f''(x) = \frac{d}{dx}(e^x - \cos x) = e^x + \sin x\]

Step 3: Set second derivative to zero

Set the second derivative equal to zero to find potential inflection points:\[e^x + \sin x = 0\]

Step 4: Approximate roots with Newton-Raphson method

To approximate roots of \(e^x + \sin x = 0\), use the Newton-Raphson formula: \[x_{n+1} = x_n - \frac{f''(x_n)}{f'''(x_n)}\]First, find \(f'''(x)\):\[f'''(x) = e^x + \cos x\]Choose an initial guess \(x_0\) within the interval \([-e, \infty)\) and apply the formula iteratively until the result stabilizes.

Step 5: Choose initial guess

Since \(-e\) is approximately \(-2.718\), choose a reasonable starting point such as \(x_0 = -2\), and apply the Newton-Raphson iteration:\[x_1 = x_0 - \frac{e^{x_0} + \sin x_0}{e^{x_0} + \cos x_0}\]

Step 6: Iterate until convergence

Apply the iteration formula repeatedly:1. Compute \(x_{n+1} = x_n - \frac{e^{x_n} + \sin x_n}{e^{x_n} + \cos x_n}\)2. Stop when \(x_n\) does not change significantly.

Key Concepts

Inflection Points

Inflection points are special points on a curve where the concavity changes. This means that the curve changes from being "concave up" (or cup-shaped) to "concave down" (or cap-shaped), or vice versa.

To find inflection points manually, it's crucial to use the second derivative of the function, denoted as \(f''(x)\). Here's a simplified way to understand it:

• Find the second derivative of the function
• Set it equal to zero \(f''(x) = 0\), and solve for \(x\)
• These \(x\) values are candidates for inflection points
• Make sure to check if the concavity changes around these points

Using these steps helps identify where inflection points are located and their corresponding \(y\) values. Remember, inflection points are where the "second derivative test" is zero and changes sign.

Derivatives

The concept of derivatives is foundational to calculus. Derivatives provide crucial information about functions, including their rates of change. The first derivative, symbolized as \(f'(x)\), tells us the slope of the tangent line to the curve at any given point. Here's why derivatives matter:

• The first derivative helps find slope and identify local maximum and minimum points
• The second derivative, \(f''(x)\), informs us about the concavity of the function, which is key in finding inflection points

In the original problem, the first derivative is \(f'(x) = e^x - \cos x\), and the second derivative is \(f''(x) = e^x + \sin x\). These derivatives were used to find and confirm the inflection points of the function \(f(x) = e^x - \sin x\).

Understanding how to calculate and apply derivatives is essential in analyzing the behavior of functions.

Graphing Functions

Graphing functions helps visualize the behavior of mathematical expressions, providing a clear picture of their slopes, curves, and critical points. When graphing a function:\[f(x) = e^x - \sin x\], it is helpful to be aware of a few things:

• The graph of the first derivative, \(f'(x)\), displays how the slope of the original function changes
• The plot of the second derivative, \(f''(x)\), can reveal where the inflection points occur

Use technology like graphing calculators or software to plot these graphs. It makes observing changes in slope or concavity more accessible, offering deeper insights into where the curve rises, falls, or shifts its bend direction. Graphs also aid in verifying analytic solutions.

Calculus Problem Solving

Calculus problem-solving involves a systematic approach to dealing with complex functions and their properties. The problem-solving process typically includes the following steps:

• Begin by differentiating the function, identifying the first and second derivatives
• Determine critical points using the first derivative and potential inflection points with the second derivative
• Apply numerical methods like the Newton-Raphson method to refine or approximate solutions

The Newton-Raphson method, specifically, is an iterative tool used for finding successively better approximations of roots, especially helpful in complex calculus problems. This method helps to hone in on exact points that purely algebraic techniques might not easily resolve.

By integrating differentiation, graphing, and numerical methods, calculus becomes a powerful tool for dissecting and understanding a wide variety of problems.