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Chapter 2: Problem 50

Determine whether \(f^{\prime}(0)\) exists. $$ f(x)=\left\\{\begin{array}{ll}{x^{2} \sin \frac{1}{x}} & {\text { if } x \neq 0} \\ {0} & {\text { if } x=0}\end{array}\right. $$

Short Answer

Expert verified
Yes, \(f'(0)\) exists and is 0.

Step by step solution

01

Understand the Definition of Derivative

The derivative of a function at a point, say at \(x = 0\), is defined as \(f'(0) = \lim_{h\to 0}\frac{f(h) - f(0)}{h}\). To find whether \(f'(0)\) exists, we need to check if this limit exists.
02

Substitute Values into the Limit Definition

Since \(f(0) = 0\), the formula becomes \(f'(0) = \lim_{h\to 0}\frac{f(h)}{h}\). For \(h eq 0\), \(f(h) = h^2 \sin \frac{1}{h}\), so \(f'(0) = \lim_{h\to 0}\frac{h^2 \sin \frac{1}{h}}{h}\).
03

Simplify the Expression

Simplifying the expression \( f'(0) = \lim_{h\to 0} h \sin \frac{1}{h}\). Since \( \sin \frac{1}{h} \) is bounded between -1 and 1, we can assert that \(-h \leq h \sin \frac{1}{h} \leq h\).
04

Apply the Squeeze Theorem

Since \(-h \leq h \sin \frac{1}{h} \leq h\), and \(\lim_{h \to 0} -h = 0\) and \(\lim_{h \to 0} h = 0\), by the Squeeze Theorem, we also have \(\lim_{h \to 0} h \sin \frac{1}{h} = 0\).
05

Conclude the Existence of the Derivative

Since \(\lim_{h \to 0} h \sin \frac{1}{h} = 0\), it follows that \(f'(0) = 0\). Therefore, \(f'(0)\) exists and is equal to 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Squeeze Theorem
The Squeeze Theorem is a fundamental concept in calculus that helps us find the limit of a function trapped between two other functions that have the same limit. In simple terms, it's like squeezing something between two fixed objects to get it to a specific point. Here鈥檚 how it works:
  • If you have two functions, say, \(g(x)\) and \(h(x)\), and they "sandwich" a third function \(f(x)\) such that \(g(x) \leq f(x) \leq h(x)\), for all x in some interval around a point (except possibly at the point itself), the Squeeze Theorem can be applied.
  • If \(\lim_{x \to a} g(x) = L\) and \(\lim_{x \to a} h(x) = L\), it follows that \(\lim_{x \to a} f(x) = L\) as well, even if \(f(x)\) is not defined at \(x = a\).
In the context of our problem, we used the Squeeze Theorem to show that \(\lim_{h \to 0} h \sin \frac{1}{h} = 0\). This is because we know \-h \leq h \sin \frac{1}{h} \leq h\ and both \(-h\) and \(h\) approach 0 as \(h\) approaches 0. Thus, the function \(h \sin \frac{1}{h}\) is squeezed to 0 as well.
Limit Definition of Derivative
The limit definition of a derivative provides the foundation for understanding and calculating derivatives. It essentially captures how a function changes as its input changes and serves as a formal definition of the rate of change of the function.
  • The derivative \(f'(a)\) of a function \(f(x)\) at a point \(x = a\) is defined as \(\lim_{h\to 0}\frac{f(a+h) - f(a)}{h}\).
  • This expression measures the slope of the tangent line to the curve \(f(x)\) at a specific point \(x = a\).
In our problem, we apply this definition at \(x = 0\) to determine the derivative exists. By substituting into the expression, we analyze the limit \(\lim_{h\to 0}\frac{f(h) - 0}{h}\), which simplifies to \(\lim_{h\to 0} h \sin \frac{1}{h}\). Hence, the derivative exists if the limit is finite, as shown by applying the Squeeze Theorem.
Piecewise Function
A piecewise function is defined differently over various intervals of its domain. It's like a rulebook with different rules depending on the situation. Piecewise functions are invaluable in modeling real-world scenarios where a function only applies under certain conditions.
  • Each "piece" of the function applies to a specific interval of the input variable \(x\).
  • In our exercise, \(f(x)\) is defined as: \(f(x) = \begin{cases} x^2 \sin \frac{1}{x}, & \text{if } x eq 0\ 0, & \text{if } x = 0 \end{cases}\).
This form of the function captures different behaviors near \(x = 0\). The special case at \(x = 0\) is necessary to ensure continuity and differentiability challenges at that specific point. Our task is to find if the derivative exists at \(x = 0\), which involves checking the behavior of the function as \(x\) approaches 0 from both sides.
Calculus Problem Solving
When tackling calculus problems, particularly those involving derivatives and limits, it鈥檚 essential to follow a logical and systematic approach. Here鈥檚 a step-by-step guide to enhance your problem-solving skills:
  • Understand the problem: Read the whole problem and understand what is being asked. In our case, it was about finding if the derivative \(f'(0)\) exists.
  • Apply definitions: Use the definitions relevant to the problem. For derivatives, apply the limit definition to understand what needs to be computed.
  • Simplify and analyze functions: Substitute values carefully and simplify the expressions whenever possible. Check if special techniques like the Squeeze Theorem can help.
  • Conclude with exactness: After calculations and simplifications, draw a firm conclusion based on the established theory; we concluded that \(f'(0) = 0\) exists.
By maintaining persistence and logical thinking, difficult calculus problems can be tackled efficiently. Understanding each concept bolsters your ability to solve them intuitively.

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Most popular questions from this chapter

$$ \begin{array}{l}{\text { If } g \text { is a twice differentiable function and } f(x)=x g\left(x^{2}\right),} \\ {\text { find } f^{\prime \prime} \text { in terms of } g, g^{\prime}, \text { and } g^{\prime \prime}}\end{array} $$

The cost, in dollars, of producing \(x\) yards of a certain fabric is $$C(x)=1200+12 x-0.1 x^{2}+0.0005 x^{3}$$ (a) Find the marginal cost function. (b) Find \(C^{\prime}(200)\) and explain its meaning. What does it predict? (c) Compare \(C^{\prime}(200)\) with the cost of manufacturing the 201 st yard of fabric.

If \(f(x)=2 x^{2}-x^{3},\) find \(f^{\prime}(x), f^{\prime \prime}(x), f^{\prime \prime \prime}(x),\) and \(f^{(4)}(x)\)Graph \(f, f^{\prime}, f^{\prime \prime},\) and \(f^{\prime \prime \prime}\) on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?

Prove, using the definition of derivative, that if \(f(x)=\cos x,\) then \(f^{\prime}(x)=-\sin x .\)

If the equation of motion of a particle is given by \(s=A \cos (\omega t+\delta),\) the particle is said to undergo simple harmonic motion. $$ \begin{array}{l}{\text { (a) Find the velocity of the particle at time } t .} \\\ {\text { (b) When is the velocity } 0 ?}\end{array} $$
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