Explanation Given: The function f ( x ) = { 3 − x x ≤ 0 3 + 1 2 x x > 0 on the interval [ − 1 , 4 ] . Calculation: The domain of the function f ( x ) = { 3 − x x ≤ 0 3 + 1 2 x x > 0 is the interval [ − 1 , 4 ] . Since, polynomials are continuous everywhere, so 3 − x is continuous in the interval ( − 1 , 0 ) and 3 + 1 2 x is continuous in the interval ( 0 , 4 ) . Now, check the continuity of the function f ( x ) at 0. To find lim x → 0 f ( x ) , it means to find the limit of the provided function as x approaches to 0 from the left that is left hand limit and as x approaches to 0 from the right that is right hand limit. Here, the function when x approaches to 0 from the left side is 3 − x . So, lim x → 0 − f ( x ) = lim x → 0 − 3 − x = 3 And, the function when x approaches to 0 from the right side is 3 + 1 2 x . So, lim x → 0 + f ( x ) = lim x → 0 + 3 + 1 2 x = 3 Here, the value of the function f ( x ) at x = 0 is 3 , which is equal to the limit of the function at x = 0 . For continuity at any point, the value of the function at that point must be equal to the limit of the function at that point. Therefore, the function is continuous at 0 . So, the function f ( x ) is continuous in the interval ( − 1 , 4 )

Bundle: Calculus of a Single Variable, 11th + WebAssign Printed Access Card for Larson/Edwards' Calculus, Multi-Term

11th Edition

ISBN: 9781337604772

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 1.4, Problem 37E

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To calculate: The continuity of the provided function f(x)={3−x x≤03+12x x>0 on the interval [−1,4].

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Chapter 1.4, Problem 36E

Chapter 1.4, Problem 38E

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Bundle: Calculus of a Single Variable, 11th + WebAssign Printed Access Card for Larson/Edwards' Calculus, Multi-Term

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The continuity of the provided function f ( x ) = { 3 − x x ≤ 0 3 + 1 2 x x > 0 on the interval [ − 1 , 4 ] .

Solution Summary: The author analyzes the continuity of the function f(x) at 0.

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To calculate: The continuity of the provided function f(x)={3−x x≤03+12x x>0 on the interval [−1,4].

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Chapter 1 Solutions

The continuity of the provided function f ( x ) = { 3 − x x ≤ 0 3 + 1 2 x x &gt; 0 on the interval [ − 1 , 4 ] .

Solution Summary: The author analyzes the continuity of the function f(x) at 0.

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To calculate: The continuity of the provided function f(x)={3−x x≤03+12x x>0 on the interval [−1,4].

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Chapter 1 Solutions

The continuity of the provided function f ( x ) = { 3 − x x ≤ 0 3 + 1 2 x x > 0 on the interval [ − 1 , 4 ] .