Math Problem Statement
Let f be the function defined by f(x) = { 3x - 2 if x < 1, ln(3x - 2) if x ≥ 1. Which of the following statements about f are true? I. lim f(x) as x approaches 1 from the left = lim f(x) as x approaches 1 from the right. II. lim f'(x) as x approaches 1 from the left = lim f'(x) as x approaches 1 from the right. III. f is differentiable at x = 1.
Solution
Let's analyze the piecewise function defined as:
3x - 2 & \text{if } x < 1 \\ \ln(3x - 2) & \text{if } x \geq 1 \end{cases}$$ We need to verify which of the following statements are true about $$f$$: --- ### I. $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$$ - For $$x \to 1^-$$, we use $$f(x) = 3x - 2$$: $$\lim_{x \to 1^-} f(x) = 3(1) - 2 = 1.$$ - For $$x \to 1^+$$, we use $$f(x) = \ln(3x - 2)$$: $$\lim_{x \to 1^+} f(x) = \ln(3(1) - 2) = \ln(1) = 0.$$ Since $$\lim_{x \to 1^-} f(x) = 1$$ and $$\lim_{x \to 1^+} f(x) = 0$$, the left-hand limit and the right-hand limit are **not equal**, so statement I is **false**. --- ### II. $$\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x)$$ Let's compute the derivatives on each side: - For $$x < 1$$, $$f(x) = 3x - 2$$, so $$f'(x) = 3$$. - For $$x \geq 1$$, $$f(x) = \ln(3x - 2)$$, so by the chain rule: $$f'(x) = \frac{1}{3x - 2} \cdot 3 = \frac{3}{3x - 2}.$$ Now, we calculate the limits of the derivatives as $$x \to 1$$: - $$\lim_{x \to 1^-} f'(x) = 3$$. - $$\lim_{x \to 1^+} f'(x) = \frac{3}{3(1) - 2} = 3$$. Since both limits are equal to 3, statement II is **true**. --- ### III. $$f$$ is differentiable at $$x = 1$$ For differentiability at $$x = 1$$, two conditions must be satisfied: 1. The function must be continuous at $$x = 1$$. 2. The left-hand and right-hand derivatives at $$x = 1$$ must be equal. We've already shown that $$\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$$, so the function is **not continuous** at $$x = 1$$, and thus, it cannot be differentiable at $$x = 1$$. Therefore, statement III is **false**. --- ### Conclusion: The only true statement is II. Hence, the correct answer is **(C) II only**. --- Would you like further clarification on any part of this solution? Here are five related questions to deepen your understanding: 1. How do you determine continuity for a piecewise function? 2. What is the relationship between continuity and differentiability? 3. Can a function be differentiable if it's not continuous? Why or why not? 4. How do you compute limits for piecewise functions? 5. Why is the chain rule used for differentiating $$\ln(3x - 2)$$? **Tip:** Always check for continuity at a point before determining differentiability!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Differentiation
Continuity
Piecewise Functions
Formulas
Limit definition
Derivative formula
Continuity condition: f(x) is continuous at x=a if lim f(x) as x approaches a exists and equals f(a).
Theorems
Continuity and differentiability
Chain rule for differentiation
Suitable Grade Level
College Calculus