Problem 74 - Calculus - AP Edition

Question

Let $$ f(x)=\left\\{\begin{array}{ll} x^{2} \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0 \end{array}\right. $$ (a) Show that \(f\) is continuous at \(x=0\) (b) Use Definition 2.2 .1 to find \(f^{\prime}(0)\) (c) Find \(f^{\prime}(x)\) for \(x \neq 0\) (d) Show that \(f^{\prime}\) is not continuous at \(x=0\)

Step-by-Step Solution

Answer

(a) f is continuous at 0. (b) f'(0) = 0. (c) f'(x) = 2x \sin(\frac{1}{x}) - \cos(\frac{1}{x}) for x≠0. (d) f' is not continuous at 0.

Step 1: Verify Continuity of f at x=0

To establish continuity at \(x=0\) for function \(f\), we must show that \(\lim_{x \to 0} f(x) = f(0)\). Calculating the limit we have \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\) for \(x eq 0\). Since \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\), it follows that \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2 \). As \(x \to 0\), both \(-x^2\) and \(x^2\) tend to 0, thus by the Squeeze Theorem, \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0 = f(0)\). Therefore, \(f\) is continuous at \(x=0\).

Step 2: Use Definition of Derivative to Find f'(0)

The derivative at \(x=0\) is given by: \(f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}\). Substituting \(f(x)\), we have \(f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h}\). Simplifying gives \(f'(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)\). Knowing \(-1 \leq \sin\left(\frac{1}{h}\right) \leq 1\), it follows that \(-h \leq h \sin\left(\frac{1}{h}\right) \leq h\) and thus, \(\lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0\) using the Squeeze Theorem. Therefore, \(f'(0) = 0\).

Step 3: Compute f'(x) for x≠0

For \(x eq 0\), use the product rule: \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\). The derivative \(f'(x) = 2x \sin\left(\frac{1}{x}\right) + x^2 \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)\) simplifies to \(f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\).

Step 4: Check Continuity of f' at x=0

From Steps 2 and 3, we know \(f'(0) = 0\) and for \(x eq 0, f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\). As \(x \to 0\), because \(\cos\left(\frac{1}{x}\right)\) oscillates between -1 and 1, \(\lim_{x \to 0} f'(x)\) does not exist due to the \(\cos\left(\frac{1}{x}\right)\) term. Thus, \(f'\) is not continuous at \(x=0\).

Key Concepts

Continuity of Functions

Continuity is a fundamental concept in calculus that describes the behavior of functions at certain points. Specifically, a function is continuous at a point if the limit of the function as it approaches that point is equal to the function's value at that point. To illustrate, consider the function given in the exercise: - For values of \(x\) not equal to 0, \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\). - At \(x = 0\), \(f(x) = 0\). To check if \(f\) is continuous at \(x = 0\), we need to verify that \(\lim_{x \to 0} f(x) = f(0)\). Using the fact that \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\), we apply the Squeeze Theorem:- Since \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\), and both \(-x^2\) and \(x^2\) approach 0 as \(x\) approaches 0, it follows that \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0\).Thus, \(f(x)\) is continuous at \(x = 0\), meaning that the function behaves predictably at this point.

Derivative at a Point

The derivative at a point provides the rate at which the function changes at that specific location. To find the derivative of a function at a particular point, we use the definition of the derivative:\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]For the function in the exercise, we are tasked with finding \(f'(0)\). Substituting the given \(f(x)\):- The expression becomes \(f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right) - 0}{h}\)- Simplifying this results in \(f'(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)\).Again, applying the fact that \(-1 \leq \sin\left(\frac{1}{h}\right) \leq 1\), we squeeze the function:- The inequality \(-h \leq h \sin\left(\frac{1}{h}\right) \leq h\) holds. - As \(h\) approaches 0, both bounds approach 0, giving \(f'(0) = 0\).This calculation shows that the instantaneous rate of change of \(f(x)\) at \(x = 0\) is zero.

Product Rule for Differentiation

The product rule is an essential tool for finding derivatives when the function is a product of two or more functions. The rule states that:\[ (uv)' = u'v + uv' \]In the context of this exercise, we apply the product rule to differentiate \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\). Here:- Let \(u = x^2\) and \(v = \sin\left(\frac{1}{x}\right)\). - Then, \(u' = 2x\) and \(v' = \cos\left(\frac{1}{x}\right) \times \left(-\frac{1}{x^2}\right)\).Applying the product rule, we find that:- \(f'(x) = 2x \sin\left(\frac{1}{x}\right) + x^2 \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right)\), which simplifies to \(f'(x) = 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)\).This expression illustrates how each part of the product is derived independently and combined to form the derivative.

Squeeze Theorem

The Squeeze Theorem is a valuable method in calculus for finding limits of functions that are difficult to evaluate directly. The principle states if a function \(g(x)\) is squeezed between two functions \(f(x)\) and \(h(x)\) near some point, and \(\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L\), then \(\lim_{x \to c} g(x) = L\) as well. In the given exercise:- The Squeeze Theorem is used twice; first, to show the continuity of \(f(x)\) at \(x = 0\), and second, to find the derivative \(f'(0)\).For continuity at \(x = 0\):- \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\). As \(x\) approaches 0, both bounds \(-x^2\) and \(x^2\) result in the same limit, helping determine \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0\).For differentiation at \(x = 0\):- \(-h \leq h \sin\left(\frac{1}{h}\right) \leq h\) gives the limit of \(h \sin\left(\frac{1}{h}\right)\) as \(h\) approaches 0, thereby calculating \(f'(0)\).The Squeeze Theorem can effectively handle functions with oscillating behaviors, which is evident in the sin component of the function, by closely bounding it.