Math Problem Statement
Let f be the piecewise-linear function defined below. Which of the following statements are true?
f(x) = { 2x - 2 for x < 3, 2x - 4 for x ≥ 3 }
I. lim(x→3) f(x) does not exist.
II. lim(x→3^+) (f(x) - f(3))/(x - 3) = 2
III. f'(3) = 2
(A) II only
(B) I and II only
(C) I, II, and III
(D) None
Solution
To analyze this question, let's go through each of the given statements concerning the piecewise-linear function :
The function is defined as:
2x - 2 & \text{for } x < 3 \\ 2x - 4 & \text{for } x \geq 3 \end{cases}$$ ### Statement I: $$\lim_{x \to 3} f(x)$$ does not exist. To check if the limit exists, we need to calculate the left-hand limit ($$x \to 3^-$$) and the right-hand limit ($$x \to 3^+$$). 1. **Left-hand limit** ($$x \to 3^-$$): $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (2x - 2) = 2(3) - 2 = 4$$ 2. **Right-hand limit** ($$x \to 3^+$$): $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x - 4) = 2(3) - 4 = 2$$ Since the left-hand limit (4) does not equal the right-hand limit (2), the limit $$\lim_{x \to 3} f(x)$$ does not exist. **Statement I is true.** ### Statement II: $$\lim_{x \to 3^+} \frac{f(x) - f(3)}{x - 3} = 2$$ To evaluate this, we need to check the derivative from the right at $$x = 3$$, which involves calculating the expression: $$\lim_{x \to 3^+} \frac{f(x) - f(3)}{x - 3}$$ First, find $$f(3)$$: $$f(3) = 2(3) - 4 = 2$$ Then, using the definition of the derivative from the right, $$\lim_{x \to 3^+} \frac{f(x) - 2}{x - 3} = \lim_{x \to 3^+} \frac{(2x - 4) - 2}{x - 3} = \lim_{x \to 3^+} \frac{2x - 6}{x - 3} = \lim_{x \to 3^+} 2 = 2$$ This limit equals 2, so **Statement II is true**. ### Statement III: $$f'(3) = 2$$ For $$f'(3)$$ to exist, $$f(x)$$ would need to be differentiable at $$x = 3$$. However, differentiability requires that the left-hand and right-hand derivatives at $$x = 3$$ are equal. 1. **Left-hand derivative** ($$x < 3$$): $$\frac{d}{dx}(2x - 2) = 2$$ 2. **Right-hand derivative** ($$x \geq 3$$): $$\frac{d}{dx}(2x - 4) = 2$$ Since both the left-hand and right-hand derivatives are equal to 2, $$f'(3) = 2$$. **Statement III is true**. ### Conclusion Since all three statements (I, II, and III) are true, the correct answer is **(C) I, II, and III**. --- Would you like a deeper explanation on any part of this solution? Here are some related questions to consider: 1. What conditions must a piecewise function meet to be differentiable at a point? 2. How do we determine the continuity of a piecewise function? 3. What role do left-hand and right-hand limits play in verifying limits at specific points? 4. How can we compute derivatives for different segments of a piecewise function? 5. How do we evaluate whether a piecewise-defined limit exists at a boundary? **Tip:** For piecewise functions, always check both left-hand and right-hand limits and derivatives separately at the boundary points.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Differentiation
Piecewise Functions
Formulas
Limit of a function as x approaches a value
Right-hand derivative definition
Theorems
Limit existence and continuity
Piecewise function differentiability
Suitable Grade Level
High School (Grades 11-12)
Related Recommendation
Testing Continuity and Differentiability of a Piecewise Function at x = 1/2
Limits and Differentiability of Piecewise Function at x=1
Differentiability and Continuity of a Piecewise Function at x = 5
Differentiability of a Piecewise Function at x = 3
Determine Continuity, Limits, and Differentiability of a Piecewise Function