Problem 58 - University Calculus: Early Transcendentals

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 parts show that the function is differentiable at \(x=0\) with \(f'(0) = 0\).

Step 1: Confirm Differentiability Condition

To show that a function \(f\) is differentiable at a point \(x_0\), we need to show that the limit \(\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}\) exists. Given \(|f(x)| \leq x^2\) for \(-1 \leq x \leq 1\), start by considering this limit at \(x = 0\).

Step 2: Evaluate Limit for Differentiability

Evaluate the limit \(\lim_{h \to 0} \frac{f(h) - f(0)}{h}\). Since \(f(0) = 0\), the expression simplifies to \(\lim_{h \to 0} \frac{f(h)}{h}\). Given \(|f(x)| \leq x^2\), we have \(-h^2 \leq f(h) \leq h^2\). Dividing by \(h\) gives \(-h \leq \frac{f(h)}{h} \leq h\).

Step 3: Apply Squeeze Theorem

Using the squeeze theorem, since \(-h \to 0\) and \(h \to 0\) as \(h \to 0\), we have \(\lim_{h \to 0} \frac{f(h)}{h} = 0\). Therefore, \(f\) is differentiable at \(x = 0\) and \(f'(0) = 0\).

Step 4: Define the Function for Part b

For part b, we have \(f(x)=\begin{cases} x^2 \sin \frac{1}{x}, & x eq 0 \ 0, & x = 0 \end{cases}\). We need to show it is differentiable at \(x=0\) and find \(f'(0)\).

Step 5: Evaluate Function Limit for Differentiability at x=0

Evaluate \(\lim_{h \to 0} \frac{f(h)}{h}\). For \(x eq 0\), \(f(h) = h^2 \sin \frac{1}{h}\). This gives \(\frac{f(h)}{h} = h \sin \frac{1}{h}\).

Step 6: Apply Squeeze Theorem for Function Limit

Since \(-1 \leq \sin \frac{1}{h} \leq 1\), it follows that \(-h \leq h \sin \frac{1}{h} \leq h\). By the squeeze theorem, \(\lim_{h \to 0} h \sin \frac{1}{h} = 0\). Thus, \(\lim_{h \to 0} \frac{f(h)}{h} = 0\).

Step 7: Conclude Differentiability and Derivative at x=0

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

Key Concepts

squeeze theorem

The Squeeze Theorem is a handy tool in calculus when you need to evaluate a limit. Imagine you have a function whose behavior or exact form you don't fully know, but it is "squeezed" between two other functions you do know, which both converge to the same limit at a certain point. Here's how it works:

• You have three functions: say, \( g(x) \), \( f(x) \), and \( h(x) \).
• These functions abide by the inequality \( g(x) \leq f(x) \leq h(x) \) for values of \( x \) near the point you're interested in.
• Both \( g(x) \) and \( h(x) \) approach the same limit, let's call it \( L \), as \( x \to a \).

Then, according to the Squeeze Theorem, \( f(x) \) must also approach \( L \) as \( x \to a \).

This principle was used multiple times in the original problem to establish the differentiability of two functions at \( x = 0 \). By recognizing that \( -h \leq \frac{f(h)}{h} \leq h \) and seeing that both \( -h \) and \( h \) converge to \( 0 \) as \( h \to 0 \), it was concluded that the limit of \( \frac{f(h)}{h} \) must also be zero, proving differentiability.

calculus limits

Understanding limits is fundamental in calculus, as it helps describe how a function behaves as its input approaches a certain point. The key idea is to see what value, if any, a function "approaches" as its input "tends towards" a particular number. Sometimes, the function may not even be defined at that specific number. Here's a quick breakdown:

• A limit looks like this: \( \lim_{x \to c} f(x) \), signifying the value \( f(x) \) approaches as \( x \) gets closer to \( c \).
• Why important? Limits help us handle indeterminate forms like \( \frac{0}{0} \) and develop key calculus concepts like continuity and differentiability.
• They require precise management because functions can behave unpredictably near specific points.

In the exercise, computing limits helped to determine the differentiability of the function at \( x = 0 \). For instance, evaluating \( \lim_{h \to 0} \frac{f(h)}{h} \) involved checking if this expression converged to a real number as \( h \to 0 \). By showing that this limit evaluates to zero, the function's behavior near zero was clarified, indicating a smooth transition without abrupt changes.

derivative evaluation

Evaluating derivatives is about determining how a function changes at a certain point. This concept, at its core, involves the notion of a rate of change or the slope of a tangent line at a given point on the function's graph.

• The derivative of a function \( f(x) \) at a point \( x_0 \) is given by the limit: \( f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \).
• For the derivative to exist at a point \( x_0 \), this limit must converge to a finite value.

The original exercise used this derivative definition at \( x = 0 \) for two functions by checking the corresponding limits.

Notice that for both functions, it was shown that \( \lim_{h \to 0} \frac{f(h)}{h} = 0 \), indicating that the derivative exists and specifically equals zero here.

• This means that at \( x = 0 \), neither function is increasing nor decreasing; they both have a flat, horizontal tangent line.

Understanding derivative evaluation provides insight into the function's behavior, such as highlighting points of stability or change, which is vital for modeling real-world scenarios and solving problems efficiently.