Math Problem Statement

Does the graph of the following curve have a tangent at the origin? Give reasons for the answer. f(x) = { x^2 sin(1/x), x ≠ 0 0, x = 0 }

Solution

To determine if the graph of the given curve f(x)f(x) has a tangent at the origin, we need to analyze the differentiability of f(x)f(x) at x=0x = 0.

Given Function:

\begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0, \\ 0 & \text{if } x = 0. \end{cases}$$ ### Condition for Tangent at the Origin: For $$f(x)$$ to have a tangent at $$x = 0$$, the derivative $$f'(0)$$ must exist, which is defined as: $$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.$$ Given that $$f(0) = 0$$, this simplifies to: $$f'(0) = \lim_{h \to 0} \frac{f(h)}{h}.$$ ### Analyzing the Limit: For $$h \neq 0$$, $$f(h) = h^2 \sin\left(\frac{1}{h}\right).$$ Thus, the limit becomes: $$f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right).$$ ### Behavior of the Limit: - Since $$|\sin\left(\frac{1}{h}\right)| \leq 1$$, we have $$|h \sin\left(\frac{1}{h}\right)| \leq |h|$$. - As $$h$$ approaches 0, $$|h|$$ also approaches 0. Therefore, the limit: $$\lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0.$$ This shows that $$f'(0) = 0$$. ### Conclusion: The limit exists, meaning the derivative exists and equals 0. Hence, **the graph of the curve does have a tangent at the origin**. The correct answer is **A. Yes, because the limit exists.** --- Would you like more details on this solution, or do you have any other questions? Here are some related questions you might consider: 1. What is the geometrical interpretation of the tangent at $$x = 0$$ for this function? 2. How would the behavior of $$\sin\left(\frac{1}{x}\right)$$ alone affect the differentiability of similar functions? 3. Can you describe a function that is continuous but not differentiable at a point? 4. What happens if the function were modified to $$x^3 \sin\left(\frac{1}{x}\right)$$ for $$x \neq 0$$? 5. What role does the limit definition of the derivative play in determining the tangent line? **Tip:** When analyzing limits involving trigonometric functions, remember that bounding the trigonometric part between -1 and 1 is often a helpful first step.

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Math Problem Analysis

Mathematical Concepts

Differentiability
Limit
Trigonometric Functions

Formulas

f'(0) = lim(h -> 0) [f(h) - f(0)] / h
h^2 sin(1/h)

Theorems

Limit Definition of the Derivative
Squeeze Theorem

Suitable Grade Level

Grades 11-12