Problem 63 - Essential Calculus

Question

Use a graphing utility to graph \(y=x \sin (1 / x)\). Show that the graph is concave downward to the right of \(x=1 / \pi\).

Step-by-Step Solution

Answer

The function \( y = x \sin(\frac{1}{x}) \) is represented by a curve fluctuating around the x-axis. The second derivative of the function for \( x > \frac{1}{\pi} \) reveals that the function is concave downward because the second derivative, \( y'' \), is less than zero.

Step 1: Plot the Function

Use your preferred graphing tool to plot the function \( y = x \sin(\frac{1}{x}) \).

Step 2: Calculate the First Derivative

Find the first derivative of the function \( y = x \sin(\frac{1}{x}) \). Using the product rule and chain rule for differentiation: \( y' = \sin(\frac{1}{x}) - \frac{x \cos(\frac{1}{x})}{x^2} = \sin(\frac{1}{x}) - \frac{\cos(\frac{1}{x})}{x} \).

Step 3: Calculate the Second Derivative

Calculate the second derivative of the function \( y = x \sin(\frac{1}{x}) \). Differentiating the first derivative \( y' \) we obtain \( y'' = -\frac{\cos(\frac{1}{x})}{x^2} + \frac{2 \sin(\frac{1}{x})}{x^3} + \frac{\sin(\frac{1}{x})}{x} \).

Step 4: Investigate the Concavity on the Interval \(x > 1/\pi\)

To show that the function is concave downward for \( x > \frac{1}{\pi} \), substitute \(x\) with values larger than \( \frac{1}{\pi} \) in the second derivative. If \( y'' < 0 \), the function is concave downward.

Key Concepts

Second Derivative Test

The second derivative test is a straightforward tool used in calculus to determine the concavity of a graph, and it helps in identifying whether a critical point is a local maximum or minimum. To use the second derivative test, you calculate the second derivative of the given function, denoted as \( f''(x) \). If you find that \( f''(x) > 0 \), the function is concave upward at that point, which generally means you have a local minimum. Conversely, if \( f''(x) < 0 \), the graph is concave downward, suggesting a local maximum. In the context of our exercise, the graph of \( y = x \sin(\frac{1}{x}) \) is said to be concave down to the right of \( x = \frac{1}{\pi} \) if the second derivative \( y'' \) is less than zero when evaluated for \( x > \frac{1}{\pi} \).

Graphing Utility

A graphing utility is an essential tool for visualizing the behavior of functions, which can be incredibly helpful when trying to grasp complex mathematical concepts. It allows you to see the shape and the key features of graphs such as intercepts, turning points, and areas of varying concavity. When working with complicated functions like \( y = x \sin(\frac{1}{x}) \), visual inspection using a graphing utility makes it much easier to understand the function's overall behavior than trying to rely solely on algebraic methods. In teaching, graphing utilities also assist students in verifying their analytical findings, adding a layer of intuition to the problem-solving process.

Product Rule

The product rule is a formula used to find the derivative of a product of two functions. It states that if you have a function \( y = u(x)\cdot v(x) \), then the derivative of \( y \), denoted as \( y' \), is given by \( y' = u'(x)\cdot v(x) + u(x)\cdot v'(x) \). This rule is crucial when dealing with the multiplication of two functions that are both dependent on \( x \). In our exercise, the product rule was used to calculate the first derivative of \( y = x \sin(\frac{1}{x}) \), with \( u(x) = x \) and \( v(x) = \sin(\frac{1}{x})\), to find \( y' \).

Chain Rule

The chain rule is another fundamental rule in differentiation used when dealing with composite functions, where one function is nested inside another. Mathematically, if \( f(x) = h(g(x)) \), then the derivative \( f'(x) \) is \( h'(g(x))\cdot g'(x) \). This rule allows you to break down the differentiation process into manageable steps when you have a function within a function. In the case of our exercise, to differentiate \( \sin(\frac{1}{x}) \), we recognize that it's a composite function where the inner function is \( g(x) = \frac{1}{x} \). Using the chain rule, we can then find the derivative of the outer sine function and multiply it by the derivative of the inner function.