Explanation Given information :. f ( x ) = x − 1 x + 3 Formula Used: Quotient Rule : d d x ( f g ) = g ⋅ f ' − f ⋅ g ' g 2 Concept Involved: Find the first derivative and second derivative of the given function. Graph f , f ' , & f ' ' In the function f graph, the interval where f < 0 {below the x -axis} is the interval where the function negative and the interval where f > 0 {above the x -axis} is the interval where the function positive. In the first derivative of function f ' graph, the interval where f ' < 0 {below the x -axis} is the interval where the function decreasing and the interval where f ' > 0 {above the x -axis} is the interval where the function increasing. Critical number is the value of x where the f ' ( x ) = 0 . In the second derivative of function f ' ' graph, the interval where f ' ' < 0 {below the x -axis} is the interval where the function concave down and the interval where f ' ' > 0 {above the x -axis} is the interval where the function concave up. Calculation: Description Steps Find the boundary points of the rational function by setting up the numerator to zero and denominator to zero x − 1 = 0 be the 1st equation x + 3 = 0 be the 2nd equation Solving the equation x − 1 = 0 by • Adding 1 on both sides x − 1 = 0 x − 1 + 1 = 0 + 1 x = 1 Solving the equation x + 3 = 0 by • Subtracting 3 on both sides x + 3 = 0 x + 3 − 3 = 0 − 3 x = − 3 List the values of x that makes the function as zero f ( x ) = 0 and f ( x ) = U n d e f i n e d x = − 3 ; x = 1 List the intervals using the boundary values of x to check where the function is positive or negative ( − ∞ , − 3 ) ; ( − 3 , 1 ) ; ( 1 , ∞ ) Pick a test point from each interval and check the sign of f ( x ) = x − 1 x + 3 Substituting the test point x = − 4 ∈ ( − ∞ , − 3 ) f ( − 4 ) = ( − 4 − 1 ) / ( − 4 + 3 ) = 5 > 0 Substituting the test point x = 0 ∈ ( − 3 , 1 ) f ( 0 ) = ( 0 − 1 ) / ( 0 + 3 ) = − 1 / 3 < 0 Substituting the test point x = 2 ∈ ( 1 , ∞ ) f ( 2 ) = ( 2 − 1 ) / ( 2 + 3 ) = 1 / 5 > 0 Sign chart of the function Analyze the sign chart and identify the intervals where the function is positive and where the function is negative Function is negative in the ( − 3 , 1 ) Function is positive in the ( − ∞ , − 3 ) ∪ ( 1 , ∞ ) Find the derivative of the function f ( x ) = x − 1 x + 3 f ' ( x ) = ( x + 3 ) ⋅ d ( x − 1 ) / d x − ( x − 1 ) ⋅ d ( x + 3 ) / d x ( x + 3 ) 2 = ( x + 3 − ( x − 1 ) ) / ( x + 3 ) 2 = ( x + 3 − x + 1 ) / ( x + 3 ) 2 = 4 / ( x + 3 ) 2 Calculation (Continued): Description Steps The critical number is the value of x that makes the first derivative f ' ( x ) = 4 ( x + 3 ) − 2 equal to zero or undefined. x + 3 = 0 Solving the equation x + 3 = 0 by • Subtracting 3 on both sides x + 3 − 3 = 0 − 3 x = − 3 List the critical number that makes first derivative as zero x = − 3 List the intervals using the critical number where the function will be increasing or decreasing ( − ∞ , − 3 ) ; ( − 3 , ∞ ) Pick a test point from each interval and check the sign of f ' ( x ) = 4 ( x + 3 ) 2 Substituting the test point x = − 4 ∈ ( − ∞ , − 3 ) , f ( − 4 ) = 4 / ( − 4 + 3 ) 2 = 4 > 0 Substituting the test point x = 4 ∈ ( 3 , ∞ ) , f ( 4 ) = 4 / ( 4 + 3 ) 2 = 4 / 49 > 0 Sign chart of the first derivative

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Limits

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Chapter 3, Problem 3SC

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To find : where the function is increasing or decreasing.

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Chapter 3, Problem 2SC

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Chapter 3 Solutions

Loose-leaf Version for Calculus

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ACP Ch. 3 - Prob. BCP Ch. 3 - Prob. CCP Ch. 3 - Prob. DCP

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where the function is increasing or decreasing.

Solution Summary: The author explains how to find the first and second derivatives of the given function by setting up the numerator and denominator to zero.

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Chapter 3 Solutions