Problem 58 - Thomas Calculus

Question

a. Let \(f(x)\) be a function satisfying \(|f(x)| \leq x^{2}\) for \(-1 \leq x \leq 1\) Show that \(f\) is differentiable at \(x=0\) and find \(f^{\prime}(0) .\) b. Show that $$ f(x)=\left\\{\begin{array}{ll}{x^{2} \sin \frac{1}{x},} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right. $$ is differentiable at \(x=0\) and find \(f^{\prime}(0)\)

Step-by-Step Solution

Answer

Both functions are differentiable at \(x=0\) with \(f'(0) = 0\).

Step 1: Analyze the Given Inequality

The function satisfies \(|f(x)| \leq x^2\) for \(-1 \leq x \leq 1\). This implies that \(-x^2 \leq f(x) \leq x^2\). We want to analyze the behavior of \(f(x)\) as \(x\) approaches 0.

Step 2: Check Differentiability at x=0

To show differentiability at \(x=0\), we need to check the limit \(\lim_{x \to 0} \frac{f(x) - f(0)}{x}\). Since \(f(0)=0\), this becomes \(\lim_{x \to 0} \frac{f(x)}{x}\). Given the bounds \(-x^2 \leq f(x) \leq x^2\), we have \(-x \leq \frac{f(x)}{x} \leq x\) for \(x eq 0\). As \(x \to 0\), both \(-x\) and \(x\) approach 0, thus by the Squeeze Theorem, the limit is 0.

Step 3: Find f'(0)

Since we have established that the limit \(\lim_{x \to 0} \frac{f(x)}{x} = 0\), it follows that \(f\) is differentiable at \(x=0\) with \(f'(0) = 0\).

Step 4: Define the Function for Part b

The function is defined as a piecewise function: \( f(x) = \begin{cases} x^2 \sin(\frac{1}{x}) & \text{if } x eq 0 \ 0 & \text{if } x = 0 \end{cases} \)

Step 5: Check Continuity for Piecewise Function

To show differentiability at \(x=0\), first confirm continuity at \(x=0\). Compute \(\lim_{x \to 0} f(x)\). Since \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\), we get \(-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2\), leading to \(\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0\), confirming continuity at \(x=0\).

Step 6: Check Differentiability of Piecewise Function at x=0

Evaluate \(\lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{x^2 \sin\left(\frac{1}{x}\right)}{x} = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right)\). Since \(|\sin\left(\frac{1}{x}\right)| \leq 1\), we have \(-x \leq x \sin\left(\frac{1}{x}\right) \leq x\), and by the Squeeze Theorem, the limit is 0.

Step 7: Find f'(0) for Piecewise Function

Thus, \(f(x)\) is differentiable at \(x=0\) and \(f'(0) = 0\).

Key Concepts

Squeeze Theorem

The Squeeze Theorem is a powerful tool in calculus that helps us find limits in cases where direct substitution fails. Put simply, it works by "squeezing" a function between two other functions whose limits are known and equal at a particular point. If the boundary functions approach the same limit, then the function squeezed between them must also approach the same limit.

In the context of this exercise, we use the Squeeze Theorem to prove the differentiability of the function at a specific point. Given the inequality \(|f(x)| \leq x^2\) and the bounds \(-x^2 \leq f(x) \leq x^2\), it allows us to show that as \(x\to 0\), both \(-x\) and \(+x\) approach the limit of zero.

• By bounding \(\frac{f(x)}{x}\) using \(-x\leq \frac{f(x)}{x}\leq x\),
• and showing \(\lim_{x \to 0} -x = \lim_{x \to 0} x = 0\),

we leverage the Squeeze Theorem to conclude that \(\lim_{x \to 0} \frac{f(x)}{x} = 0\).

Once this limit is established, it follows that the derivative \(f'(0) = 0\), illustrating usability of this theorem in determining limits and differentiability.

Piecewise Functions

Piecewise functions are a key concept when working with functions defined by different expressions based on the input values. They break the domain into "pieces," where each piece has its own specific rule or expression.

This is particularly useful for capturing behaviors that change dramatically at certain points, such as when defining functions like \(f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x eq 0 \ 0 & \text{if } x = 0 \end{cases}\). Each piece of this function handles behavior for \(x \eq 0\) and \(x = 0\) individually.

When dealing with piecewise functions, identifying continuity at the junctions of their pieces is crucial:

• Check if the limit as \(x \ o 0\) from either direction matches the value at \(x = 0\).
• If both are equal, the function is continuous at that point.

For example, in our case, we verify that the limit of \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\) as \(x \to 0\) is 0, justifying the continuity at \(x = 0\). Thus, we ascertain differentiability by confirming continuity as a prior check.

Limits and Continuity

Limits and continuity form the bedrock of calculus, providing a foundation for analyzing and understanding function behavior as approaches occur. A limit describes the value that a function approaches as the input approaches a certain point, while continuity ensures a function maintains the same value at that point as its limit.

To show that a function is differentiable at a point, it must first be continuous there:

• Start by computing the limit of \(f(x)\) as \(x\) approaches the point in question, say \(x = 0\), to confirm its value matches \(f(0)\).
• Once continuity is established, evaluate differentiability by checking whether the limit of the difference quotient exists.

In practice, for the function \(f(x) = x^2 \sin\left(\frac{1}{x}\right)\),\(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\) yields \(-x^2 \leq x^2 \sin\frac{1}{x} \leq x^2\), squeezing the output closer to zero as \(x \to 0\).

Handling limits in this manner confirms continuity, providing a clear path to assessing differentiability and accurately finding derivatives at complex points.