Suppose that the piecewise defined function f is defined by 4 - 2x, x≤2 f(x) = { +² +5x-14, x>2 Determine which of the following statements are true. Select the correct answer below: Of is neither differentiable nor continuous at x = 2. Of is continuous but not differentiable at x = 2. Of is both differentiable and continuous at x = = 2.
Suppose that the piecewise defined function f is defined by 4 - 2x, x≤2 f(x) = { +² +5x-14, x>2 Determine which of the following statements are true. Select the correct answer below: Of is neither differentiable nor continuous at x = 2. Of is continuous but not differentiable at x = 2. Of is both differentiable and continuous at x = = 2.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Piecewise Defined Function Analysis**
**Problem Statement:**
Suppose that the piecewise defined function \( f \) is defined by
\[ f(x) =
\begin{cases}
4 - 2x & \text{, } x \leq 2 \\
x^2 + 5x - 14 & \text{, } x > 2
\end{cases}
\]
Determine which of the following statements are true.
**Select the correct answer below:**
- \( \circ \) \( f \) is neither differentiable nor continuous at \( x = 2 \).
- \( \circ \) \( f \) is continuous but not differentiable at \( x = 2 \).
- \( \circ \) \( f \) is both differentiable and continuous at \( x = 2 \).
**Explanation:**
* To determine continuity at \( x = 2 \), we need to check if \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \).
* To determine differentiability at \( x = 2 \), we must see if the left-hand derivative (\(\lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h}\)) equals the right-hand derivative (\(\lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}\)).
This information helps students understand discontinuities and differentiability for piecewise functions at specific points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5ea397a-632d-4020-8701-e2de67ac8a17%2F2b67e6c3-8355-4ad6-b410-ce0d1605e80b%2Ftb2pfy_processed.png&w=3840&q=75)
Transcribed Image Text:**Piecewise Defined Function Analysis**
**Problem Statement:**
Suppose that the piecewise defined function \( f \) is defined by
\[ f(x) =
\begin{cases}
4 - 2x & \text{, } x \leq 2 \\
x^2 + 5x - 14 & \text{, } x > 2
\end{cases}
\]
Determine which of the following statements are true.
**Select the correct answer below:**
- \( \circ \) \( f \) is neither differentiable nor continuous at \( x = 2 \).
- \( \circ \) \( f \) is continuous but not differentiable at \( x = 2 \).
- \( \circ \) \( f \) is both differentiable and continuous at \( x = 2 \).
**Explanation:**
* To determine continuity at \( x = 2 \), we need to check if \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \).
* To determine differentiability at \( x = 2 \), we must see if the left-hand derivative (\(\lim_{h \to 0^-} \frac{f(2+h) - f(2)}{h}\)) equals the right-hand derivative (\(\lim_{h \to 0^+} \frac{f(2+h) - f(2)}{h}\)).
This information helps students understand discontinuities and differentiability for piecewise functions at specific points.
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