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8. Let f(x) = x² if x ≤ 0 and f(x) = x + 1 if a > 0. At what points f: RR continuous? Here R is the set of real numbers. Justify your answer.

8. Let f(x) = x² if x ≤ 0 and f(x) = x + 1 if a > 0. At what points f: RR continuous? Here R is the set of real numbers. Justify your answer.

Advanced Engineering Mathematics
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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**Problem Statement** Let \( f(x) = -x^2 \) if \( x \leq 0 \) and \( f(x) = x + 1 \) if \( x > 0 \). At what points is the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) continuous? Here, \(\mathbb{R}\) represents the set of real numbers. Justify your answer. **Analysis** The function \( f(x) \) is defined piecewise: 1. For \( x \leq 0 \), the function is \( f(x) = -x^2 \), a parabola opening downwards with vertex at the origin. 2. For \( x > 0 \), the function is \( f(x) = x + 1 \), a linear function with a slope of 1 and a y-intercept at 1. To determine the points of continuity, we need to check if the function is continuous at the point \( x = 0 \) where the definition changes, as well as its general continuity on the intervals where each piece is defined. **Justification** - **Continuity for \( x < 0 \)**: Since \( f(x) = -x^2 \) is a polynomial, it is continuous for all \( x \leq 0 \). - **Continuity for \( x > 0 \)**: The function \( f(x) = x + 1 \) is also a polynomial, hence continuous for all \( x > 0 \). - **Continuity at \( x = 0 \)**: To check the continuity at \( x = 0 \): - **Limit from the left**: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x^2 = 0 \). - **Limit from the right**: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1 \). - **Function value**: \( f(0) = -0^2 = 0 \). Since the left-hand limit at 0 is 0 and the right-hand limit is 1, and these limits do not match, the function is not continuous at \( x =
Transcribed Image Text:**Problem Statement** Let \( f(x) = -x^2 \) if \( x \leq 0 \) and \( f(x) = x + 1 \) if \( x > 0 \). At what points is the function \( f : \mathbb{R} \rightarrow \mathbb{R} \) continuous? Here, \(\mathbb{R}\) represents the set of real numbers. Justify your answer. **Analysis** The function \( f(x) \) is defined piecewise: 1. For \( x \leq 0 \), the function is \( f(x) = -x^2 \), a parabola opening downwards with vertex at the origin. 2. For \( x > 0 \), the function is \( f(x) = x + 1 \), a linear function with a slope of 1 and a y-intercept at 1. To determine the points of continuity, we need to check if the function is continuous at the point \( x = 0 \) where the definition changes, as well as its general continuity on the intervals where each piece is defined. **Justification** - **Continuity for \( x < 0 \)**: Since \( f(x) = -x^2 \) is a polynomial, it is continuous for all \( x \leq 0 \). - **Continuity for \( x > 0 \)**: The function \( f(x) = x + 1 \) is also a polynomial, hence continuous for all \( x > 0 \). - **Continuity at \( x = 0 \)**: To check the continuity at \( x = 0 \): - **Limit from the left**: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} -x^2 = 0 \). - **Limit from the right**: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 1 \). - **Function value**: \( f(0) = -0^2 = 0 \). Since the left-hand limit at 0 is 0 and the right-hand limit is 1, and these limits do not match, the function is not continuous at \( x =
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