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•. Sketch the graph of a continous function & f(2)=1 f'(2)=0, the point (2,1) is 1 1 1 1 2 I is undefined at x=3 e) <0 on (-0,3) ~ (3,00) + " (*) <0 on (-∞, 3). {"(x) > 0 on (3, 0) so that a local minimum of

•. Sketch the graph of a continous function & f(2)=1 f'(2)=0, the point (2,1) is 1 1 1 1 2 I is undefined at x=3 e) <0 on (-0,3) ~ (3,00) + " (*) <0 on (-∞, 3). {"(x) > 0 on (3, 0) so that a local minimum of

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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please answer question 1 ans 2

### Problem 2: Sketch the Graph of a Continuous Function \( f \) 1. The conditions for the function \( f \) are as follows: - \( f(2) = 1 \) - \( f'(2) = 0 \), meaning the point \( (2, 1) \) is a local minimum of \( f \). 2. Further conditions: - \( f \) is undefined at \( x = 3 \). - The first derivative \( f'(x) < 0 \) on \( (-\infty, 3) \cup (3, \infty) \). - The second derivative \( f''(x) < 0 \) on \( (-\infty, 3) \). - The second derivative \( f''(x) > 0 \) on \( (3, \infty) \). ### Explanation: This problem asks us to sketch the graph of a continuous function \( f \) given specific conditions concerning its values, derivatives, and points of discontinuity. #### Detailed Conditions: 1. **For \( x = 2 \):** - The function passes through the point \( (2,1) \), i.e., \( f(2) = 1 \). - The function has a zero slope at \( x = 2 \), \( f'(2) = 0 \), meaning it is a critical point. - The point \( (2, 1) \) is a local minimum. 2. **Around \( x = 3 \):** - The function is not defined at \( x = 3 \). - \( f'(x) \) is negative to the left and right of \( x = 3 \), implying the function is decreasing on these intervals. - The concavity of the function changes around \( x = 3 \) as indicated by \( f''(x) \): - \( f''(x) < 0 \) on \( (-\infty, 3) \), indicating concave down (i.e., the graph bends downwards). - \( f''(x) > 0 \) on \( (3, \infty) \), indicating concave up (i.e., the graph bends upwards). These conditions will help in sketching the graph precisely, noting the abrupt change in behavior at
Transcribed Image Text:### Problem 2: Sketch the Graph of a Continuous Function \( f \) 1. The conditions for the function \( f \) are as follows: - \( f(2) = 1 \) - \( f'(2) = 0 \), meaning the point \( (2, 1) \) is a local minimum of \( f \). 2. Further conditions: - \( f \) is undefined at \( x = 3 \). - The first derivative \( f'(x) < 0 \) on \( (-\infty, 3) \cup (3, \infty) \). - The second derivative \( f''(x) < 0 \) on \( (-\infty, 3) \). - The second derivative \( f''(x) > 0 \) on \( (3, \infty) \). ### Explanation: This problem asks us to sketch the graph of a continuous function \( f \) given specific conditions concerning its values, derivatives, and points of discontinuity. #### Detailed Conditions: 1. **For \( x = 2 \):** - The function passes through the point \( (2,1) \), i.e., \( f(2) = 1 \). - The function has a zero slope at \( x = 2 \), \( f'(2) = 0 \), meaning it is a critical point. - The point \( (2, 1) \) is a local minimum. 2. **Around \( x = 3 \):** - The function is not defined at \( x = 3 \). - \( f'(x) \) is negative to the left and right of \( x = 3 \), implying the function is decreasing on these intervals. - The concavity of the function changes around \( x = 3 \) as indicated by \( f''(x) \): - \( f''(x) < 0 \) on \( (-\infty, 3) \), indicating concave down (i.e., the graph bends downwards). - \( f''(x) > 0 \) on \( (3, \infty) \), indicating concave up (i.e., the graph bends upwards). These conditions will help in sketching the graph precisely, noting the abrupt change in behavior at
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