Problem 71 - Single Variable Calculus

Question

Let $$ f(x)=\left\\{\begin{array}{ll} \left(2-\sin \frac{1}{x}\right)|x| & \text { if } x \neq 0 \\ 0 & \text { if } x=0 \end{array}\right. $$ a. Plot the graph of \(f\). Use zooM to obtain successive magnifications of the graph in the neighborhood of the origin. Can you see that \(f\) has a relative minimum at 0 but is not monotonic to the left or to the right of \(x=0 ?\) b. Prove the observation made in part (a). Hint: For \(x>0\), show that \(f^{\prime}(x)>0\) if \(x=1 /(2 n \pi)\) and \(f^{\prime}(x)<0\) if \(x=1 /((2 n+1) \pi)\)

Step-by-Step Solution

Answer

In summary, the graph of the function \(f(x)\) indeed has a relative minimum at \(x=0\). The function is not monotonic to the left or to the right of \(x=0\) as its derivative \(f'(x)\) changes signs around \(x=0\). Specifically, \(f'(x)>0\) if \(x=\frac{1}{2n\pi}\) and \(f'(x)<0\) if \(x=\frac{1}{(2n+1)\pi}\), where n is an integer.

Step 1: a. Plot the graph of f(x) and zoom in near the origin

Firstly, plot the graph of f(x) using a graphing calculator or a plotting software. Observe the graph behavior near the origin (x = 0). Then, use the zoom feature to magnify the graph in the neighborhood of the origin. By observing the magnified graph, the function appears to have a relative minimum at x = 0. However, the function is not monotonic on either side of x = 0.

Step 2: b. Prove the observation

To prove the observation, we will do the following steps: 1. Find the first derivative of f(x) with respect to x for x > 0. 2. Show that \(f'(x) > 0\) if \(x = \frac{1}{2n\pi}\) and \(f'(x) < 0\) if \(x = \frac{1}{(2n + 1)\pi}\). Step 1: Find the first derivative of f(x) with respect to x for x > 0. Since the function f(x) is defined differently for x ≠ 0 and x = 0, we only need to find the derivative of \(f(x) = (2 - \sin(\frac{1}{x}))|x|\) for x > 0. Using the product and chain rules, we get: \[f'(x) = (2-\sin(\frac{1}{x}))\frac{d(|x|)}{dx} - |x|\frac{d(\sin(\frac{1}{x}))}{dx}\] For x > 0, \(|x| = x\), and \[\frac{d(\sin(\frac{1}{x}))}{dx} = \cos(\frac{1}{x})(-\frac{1}{x^2})\] So, \[f'(x) = (2 - \sin(\frac{1}{x})) - (-\frac{x\cos(\frac{1}{x})}{x^2})\] Step 2: Show that \(f'(x) > 0\) if \(x = \frac{1}{2n\pi}\) and \(f'(x) < 0\) if \(x = \frac{1}{(2n + 1)\pi}\). If \(x = \frac{1}{2n\pi}\), then \(\frac{1}{x} = 2n\pi\) and \(\sin(\frac{1}{x}) = \sin(2n\pi) = 0\). Therefore, \[f'(x) = 2 - 0 + \frac{\cos(2n\pi)}{2n\pi} > 0\] If \(x = \frac{1}{(2n + 1)\pi}\), then \(\frac{1}{x} = (2n + 1)\pi\) and \(\sin(\frac{1}{x}) = \sin((2n + 1)\pi) = 0\). Therefore, \[f'(x) = 2 - 0 - \frac{\cos((2n + 1)\pi)}{(2n + 1)\pi} < 0\] Hence, the function f(x) has a relative minimum at x = 0, as its derivative changes sign around x = 0, but is not monotonic to the left or to the right of x = 0 as there are alternating intervals with positive and negative derivatives.

Key Concepts

Derivative

Understanding derivatives is essential in calculus as they describe how a function changes at any point. Mathematically, the derivative of function \(f(x)\) at a point \(x\) is defined as the limit of the average rate of change of the function as the interval shrinks to zero.
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\]Using derivatives, we can determine the function's behavior, like finding when it is increasing or decreasing. In our exercise, the derivative \(f'(x)\) gives insight into whether \(f(x)\) is going up or down as \(x\) changes. For example, if \(f'(x) > 0\) for some interval, \(f(x)\) is increasing over that interval. Similarly, \(f'(x) < 0\) indicates a decreasing interval.

Relative Minimum

A relative minimum point on a graph is where the function has a lower value than all nearby points. In other words, it's a "valley" in the graph of the function. Identifying relative minima is crucial as it helps understand the function's behavior.
For a function to have a relative minimum at a point \(x = a\), the following conditions are typically met:

• The first derivative \(f'(a) = 0\) or \(f'(a)\) changes sign around \(a\).
• The second derivative \(f''(a) > 0\) confirms the minimum is indeed a relative minimum if the derivative changes sign.

In our specific exercise, even though \(f(0) = 0\) is a relative minimum on the graph of \(f(x)\), the function isn't necessarily monotonic, meaning it doesn't consistently increase or decrease around \(x=0\). This is confirmed by how the derivative changes sign around this point.

Monotonicity

Monotonicity refers to the specific intervals in which a function consistently increases or decreases. A function is monotonic if it is either entirely non-increasing or non-decreasing over an interval.
This property is essential for analyzing graph behavior and expectations. For instance, if a function is non-decreasing across some interval, its derivative will be non-negative within that interval.
In our exercise, despite the relative minimum at \(x = 0\), the function \(f(x)\) is noted not to be monotonic around \(x = 0\). The alternate changes in the derivative—positive at some points and negative at others—demonstrate that the function increases and decreases on either side of \(0\), thus lacking monotonicity.

Graphing

Graphing is an instrumental method in calculus for visualizing functions and understanding their behavior over different domains. Plotting a graph provides insights into key properties like continuity, extrema (such as minima and maxima), and monotonicity.
With graphing software, or even by hand using a calculator, one can approximate the shape and nature of any function. By observing a graph, we can infer the nature of the infinitesimal changes occurring at any point, which might not be easily visible only from algebraic expressions.
For example, graphing \(f(x)\) from our exercise visually confirms the relative minimum at \(x = 0\) and the lack of monotonicity around that point. This visual method complements derivative calculations by adding another layer of understanding to mathematical analysis.