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a 3. Sketch the graph of +₁ f(₂)=1₁ continous function & $²(2)=0, the point (2,1) is 2 £ is undefined at *=3 11(e) <0 on (-∞0,3) ~ (3,00) +" (20) <0 on (-∞, 3) {"(x) >0 on (3,00). so that a local minimum of

a 3. Sketch the graph of +₁ f(₂)=1₁ continous function & $²(2)=0, the point (2,1) is 2 £ is undefined at *=3 11(e) <0 on (-∞0,3) ~ (3,00) +" (20) <0 on (-∞, 3) {"(x) >0 on (3,00). 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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### Sketching the Graph of a Continuous Function \( f \) To sketch the graph of a continuous function \( f \) that satisfies the given conditions, follow the steps and properties listed below: #### Properties and Conditions: 1. \( f(2) = 1 \) - The function passes through the point \( (2, 1) \). 2. \( f'(2) = 0 \) - The slope of the tangent at \( x = 2 \) is zero, indicating that \( (2, 1) \) is a local extremum. - Given that it is specified as a local minimum, the curve of \( f(x) \) should be concave up at \( (2, 1) \), meaning \( f''(2) > 0 \). 3. \( f \) is undefined at \( x = 3 \) - There is a discontinuity (e.g., a vertical asymptote or a hole) at \( x = 3 \). 4. \( f'(x) < 0 \) on \( (-\infty, 3) \cup (3, \infty) \) - The first derivative \( f'(x) \) is negative on these intervals, which means the function \( f(x) \) is decreasing before and after \( x = 3 \). 5. \( f''(x) < 0 \) on \( (-\infty, 3) \) - The second derivative \( f''(x) \) is negative on this interval, indicating that the function \( f(x) \) is concave down before \( x = 3 \). 6. \( f''(x) > 0 \) on \( (3, \infty) \) - The second derivative \( f''(x) \) is positive on this interval, indicating that the function \( f(x) \) is concave up after \( x = 3 \). #### Steps to Sketch the Graph: 1. **Plot the Point and Minimum:** - Plot the point \( (2, 1) \). - Since \( f'(2) = 0 \) and it’s a local minimum, draw the graph concave up around \( (2, 1) \). 2. **Discontinuity at \( x = 3 \):** -
Transcribed Image Text:### Sketching the Graph of a Continuous Function \( f \) To sketch the graph of a continuous function \( f \) that satisfies the given conditions, follow the steps and properties listed below: #### Properties and Conditions: 1. \( f(2) = 1 \) - The function passes through the point \( (2, 1) \). 2. \( f'(2) = 0 \) - The slope of the tangent at \( x = 2 \) is zero, indicating that \( (2, 1) \) is a local extremum. - Given that it is specified as a local minimum, the curve of \( f(x) \) should be concave up at \( (2, 1) \), meaning \( f''(2) > 0 \). 3. \( f \) is undefined at \( x = 3 \) - There is a discontinuity (e.g., a vertical asymptote or a hole) at \( x = 3 \). 4. \( f'(x) < 0 \) on \( (-\infty, 3) \cup (3, \infty) \) - The first derivative \( f'(x) \) is negative on these intervals, which means the function \( f(x) \) is decreasing before and after \( x = 3 \). 5. \( f''(x) < 0 \) on \( (-\infty, 3) \) - The second derivative \( f''(x) \) is negative on this interval, indicating that the function \( f(x) \) is concave down before \( x = 3 \). 6. \( f''(x) > 0 \) on \( (3, \infty) \) - The second derivative \( f''(x) \) is positive on this interval, indicating that the function \( f(x) \) is concave up after \( x = 3 \). #### Steps to Sketch the Graph: 1. **Plot the Point and Minimum:** - Plot the point \( (2, 1) \). - Since \( f'(2) = 0 \) and it’s a local minimum, draw the graph concave up around \( (2, 1) \). 2. **Discontinuity at \( x = 3 \):** -
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