a. Explanation Concept Used: In this we have to use the concept of limits. Calculation: We note that x − 2 is negative when x < 2 , and positive when x > 2 .So to write the function without the absolute value ,it becomes g ( x ) = { x 2 + x − 6 − ( x − 2 ) , i f x < 2 ; x 2 + x − 6 x − 2 , i f x > 2 Putting the negative there makes the denominator positive when x > 2 ,so the expression can have the same value as the original with the absolute value. (i) For x − > 2 + we use the x > 2 version of g(x) lim x → 2 + x 2 + x − 6 x − 2 = lim x → 2 + ( x + 3 ) ( x − 2 ) x − 2 = b. To determine To find whether the lim x → 2 g ( x ) exists. c. To determine To draw the graph of g.

15th Edition

ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Chapter 2.4, Problem 37E

To determine

To find the limit of limx→2+g(x) and limx→2−g(x) .

To determine

To find whether the limx→2g(x) exists.

To determine

To draw the graph of g.

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Chapter 2.4, Problem 36E

Chapter 2.4, Problem 38E

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Chapter 2 Solutions

Biocalculus

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To find the limit of lim x → 2 + g ( x ) and lim x → 2 − g ( x ) .

Solution Summary: The author calculates the limit of undersetx to 2+mathrmlimg(x), and the denominator is positive.

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To find the limit of limx→2+g(x) and limx→2−g(x) .

To find whether the limx→2g(x) exists.

To draw the graph of g.

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Chapter 2 Solutions