Problem 32 - Single Variable Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll} x \sin \frac{\pi}{x} & \text { if } x>0 \\ 0 & \text { if } x=0 \end{array}\right. $$ Use Rolle's Theorem to prove that \(f\) has infinitely many critical numbers in the interval \((0,1)\). Plot the graph of \(f\) using the viewing window \([0,1] \times[-1,1]\).

Step-by-Step Solution

Answer

In summary, to prove that the function \(f(x)=\left\{\begin{array}{ll} x \sin \frac{\pi}{x} & \text { if } x>0 \\\ 0 & \text { if } x=0 \end{array}\right.\) has infinitely many critical numbers in the interval (0,1) using Rolle's theorem, we first confirm that the function is continuous and differentiable. Then, we show that for any \(n \in \mathbb{N}\), there exists a \(c_n\) in the open interval \(0 < c_n < \frac{1}{n}\) such that \(f'(c_n) = 0\). Hence, there are infinitely many critical numbers in the given interval. To plot the graph within the viewing window \(x \in [0,1]\), \(y \in [-1,1]\), use any graphing calculator or software like Desmos, Wolfram Alpha, or Geogebra.

Step 1: Examine the continuity and differentiability of the function

Firstly, we need to check whether the function is continuous and differentiable on the interval (0,1). The function \(f(x) = x\sin(\frac{\pi}{x})\) is continuous in the given interval as both \(x\) and \(\sin(\frac{\pi}{x})\) are continuous. Now let's check for differentiability. We will need to find the derivative of our function \(f(x)\).

Step 2: Calculate the derivative of the function (if x ≠ 0)

We can differentiate \(f(x)=x\sin(\frac{\pi}{x})\) using the product rule, which states: If \(u(x)\) and \(v(x)\) are differentiable functions, then \(\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)\) Let \(u(x) = x\) and \(v(x) = sin(\frac{\pi}{x})\). Then, the derivatives of these functions are: \(u'(x) = 1\) \(v'(x) = \frac{-\pi\cos(\frac{\pi}{x})}{x^2}\) Now, applying the product rule, we get: \(f'(x) = 1 \cdot \sin(\frac{\pi}{x}) + x \cdot \frac{-\pi\cos(\frac{\pi}{x})}{x^2}\) After simplifying, we obtain: \(f'(x) = \sin(\frac{\pi}{x}) - \frac{\pi\cos(\frac{\pi}{x})}{x}\)

Step 3: Apply Rolle's Theorem

For \(x > 0\), we apply Rolle's theorem on the interval \((0,1)\). We will try to find the \(n^{th}\) critical number as we want to prove that there are infinitely many critical numbers. Consider the function \(f(x)\) on the interval [0, \(\frac{1}{n}\)] where \(n \in \mathbb{N}\). According to Rolle's theorem, if \(f(0) = f(\frac{1}{n})\), then there exists at least one number \(c_n\) in the open interval \(0 < c_n < \frac{1}{n}\) such that \(f'(c_n) = 0\). As \(f(0) = 0\), we need to find \(f(\frac{1}{n})\): \(f\left(\frac{1}{n}\right)=\frac{1}{n}\sin(\pi n)\) Since, \(n \in \mathbb{N}\), \(\sin(\pi n) = 0\), thus: \(f\left(\frac{1}{n}\right)=\frac{1}{n}\cdot 0 = 0\) Now, we have \(f(\frac{1}{n}) = 0 = f(0)\). By Rolle's theorem, there must exist at least one \(c_n\) in the open interval \(0 < c_n < \frac{1}{n}\) such that \(f'(c_n) = 0\). Since \(n\) can be any natural number, this implies there are infinitely many critical numbers in the interval (0,1) for the function \(f(x)\).

Step 4: Plot the graph

To plot the graph of \(f(x)\) using the viewing window \(x \in [0,1]\), \(y \in [-1,1]\), you can use any graphing calculator or graphing software (like Desmos, Wolfram Alpha, or Geogebra). Simply input the function \(f(x) = x\sin(\frac{\pi}{x})\) and set the viewing window to the given interval. You will be able to see how the function behaves in the interval and observe the points at which the slope of the function is 0, indicating critical numbers.

Key Concepts

Critical Numbers

Critical numbers are values in the domain of a function where the derivative is either zero or undefined. These points are crucial because they can indicate where the function has a maximum, minimum, or even a point of inflection.
For a function like our given piecewise function, identifying these points involves checking where the derivative is zero or does not exist in the interval \(0, 1\).
In the case of our exercise, when applying Rolle's Theorem, we found that critical numbers exist in the sub-intervals like \(0, \frac{1}{n}\).

• Because as \(n\) increases, this pattern repeats leading to infinite possible values for critical numbers in our interval.
• This not only confirms the behavior of the function but also illustrates its oscillatory nature.

Understanding critical numbers help us to scrutinize the function behavior fully and how it changes across its domain.

Derivative

The derivative of a function represents the rate of change of the function with respect to its variable. In simpler terms, it tells us how steep the function is at any given point.
When finding the derivative for the function \(f(x) = x \sin(\frac{\pi}{x})\), we use differentiation techniques such as the product rule to determine the change.

• The derivative provides vital information about the function's behavior, especially when finding critical numbers.
• A solution to \(f'(x) = 0\) often leads us to the critical numbers, playing a key part in applying Rolle's Theorem.

Regularly assessing the derivative is necessary as it allows for a deeper inspection into how quickly and in what direction the function is changing.

Product Rule

The product rule is a common method used in calculus when differentiating products of two functions. This rule is key when you have functions multiplied together, like in our current function.
For our exercise, the function \(f(x) = x \sin(\frac{\pi}{x})\) was differentiated using this rule as follows:

• If \(u(x)\) and \(v(x)\) are functions, then the derivative is \(uv' + vu'\).
• By setting \(u(x) = x\) and \(v(x)= \sin(\frac{\pi}{x})\), the derivative becomes \(\sin(\frac{\pi}{x}) - \frac{\pi \cos(\frac{\pi}{x})}{x}\).

Understanding the product rule is extremely beneficial for recognizing the probable changes and variations in a product function, leading to easier discovery of critical points.

Continuity and Differentiability

In calculus, continuity means a function doesn't have any abrupt changes or gaps in its value, while differentiability is about a function having a derivative at every point in its domain.
For \(f(x) = x \sin(\frac{\pi}{x})\), it’s essential to first establish that the function is continuous over the open interval \(0, 1\).

• Continuity can be checked by ensuring no jumps or breaks in this interval.
• Differentiability, on the other hand, requires finding the derivative and checking that it exists across all points except perhaps points of concern like x=0.

These properties are foundational when applying Rolle's Theorem, as they ensure the prerequisites that make finding critical numbers possible. They assure us of predictive behaviors in mathematical models, which are crucial when evaluating real-world phenomena or abstract concepts.