Explanation Given information : f ( x ) = x x 2 + 1 Formula Used: Power rule of derivative: If f ( x ) = k x n then f ' ( x ) = k ⋅ n x n − 1 Quotient Rule : d d x ( f g ) = g ⋅ f ' − f ⋅ g ' g 2 Concept Involved: Step 1: The roots of the function are the values of x that makes the function equal to zero. Step 2: Find the first derivative of the function. Step 3: Critical numbers are values of x that makes the first derivative a zero or undefined. Set up equation to find the critical numbers ‘c’. Step 4: Write all the intervals where the function will be increasing or decreasing using the critical number. Step 5: Pick a test point ‘ x 0 ’ from each interval, substitute it in the first derivative; If f ' ( x 0 ) < 0 , then function is decreasing in that interval & If f ' ( x 0 ) > 0 , then function is increasing in that interval. Step 6: Based on the below cases, identify whether the critical number that we found is local maximum or local minimum. If to the left of critical number function is decreasing and to the right function is increasing, then it means at c we have local minimum given by ( c , f ( c ) ) If to the left of critical number function is increasing and to the right function decreasing, then it means at c we have local maximum. given by ( c , f ( c ) ) Global extrema are the largest and smallest values that a function takes on over its entire domain Step 7: Find the second derivative of the function. Step 8: Find the values of x that makes the f ' ' ( x ) = 0 . Step 9: Write all the intervals where the function will be concave up or concave down using the values of x that makes the equation f ' ' ( x ) = 0 TRUE. Step 10: Pick a test point ‘ x 1 ’ from each interval, substitute it in the first derivative; If f ' ' ( x 1 ) < 0 , then function is concave down in that interval & If f ' ' ( x 1 ) > 0 , then function is concave up in that interval. Step 11: Inflection points are the values of x where the function changes concavity. Calculation: Description Steps Find the zeros of the function that makes f ( x ) = 0 . For a rational function we need to set the numerator to zero x = 0 List the values of x that makes the function as zero f ( x ) = 0 x = 0 Find the derivative of the function f ( x ) = x x 2 + 1 f ' ( x ) = ( x 2 + 1 ) ⋅ d ( x ) / d x − ( x ) ⋅ d ( x 2 + 1 ) / d x ( x 2 + 1 ) 2 = ( ( x 2 + 1 ) ( 1 ) − ( x ) ( 2 x ) ) / ( x 2 + 1 ) 2 = ( x 2 + 1 − 2 x 2 ) / ( x 2 + 1 ) 2 = ( 1 − x 2 ) / ( x 2 + 1 ) 2 Calculation (Continued): Description Steps The critical number is the value of x that makes the first derivative f ' ( x ) = ( 1 − x 2 ) / ( x 2 + 1 ) 2 equal to zero or undefined. Setting up numerator and denominator to zero 1 − x 2 = 0 be the 1 s t equation ( x 2 + 1 ) 2 = 0 be the 2 n d equation Solving the equation 1 − x 2 = 0 by • Subtracting 1 on both sides • Dividing by -1 on both sides • Taking square root on both sides 1 − x 2 − 1 = 0 − 1 − x 2 = − 1 − x 2 / − 1 = − 1 / − 1 x 2 = 1 x 2 = 1 x = ± 1 Solving the equation ( x 2 + 1 ) 2 = 0 by There is no real value of xthat makes the equation TRUE List the critical number that makes first derivative as zero x = − 1 ; x = 1 List the intervals using the critical number where the function will be increasing or decreasing ( − ∞ , − 1 ) ; ( − 1 , 1 ) ; ( 1 , ∞ ) Pick a test point from each interval and check the sign of f ' ( x ) = ( 1 − x 2 ) / ( x 2 + 1 ) 2 Substituting the test point x = − 2 ∈ ( − ∞ , − 1 ) , f ' ( − 2 ) = ( 1 − ( − 2 ) 2 ) / ( ( − 2 ) 2 + 1 ) 2 = − 0.12 < 0 Substituting the test point x = 0 ∈ ( − 1 , 1 ) , f ' ( 0 ) = ( 1 − ( 0 ) 2 ) / ( ( 0 ) 2 + 1 ) 2 = 1 > 0 Substituting the test point x = 2 ∈ ( 1 , ∞ ) , f ' ( 2 ) = ( 1 − ( 2 ) 2 ) / ( ( 2 ) 2 + 1 ) 2 = − 0.12 < 0 Analyzing the left and right of critical number x = − 1 Since f ' ( x ) < 0 to left of x = − 1 and f ' ( x ) > 0 to right of x = − 1 ; At x = − 1 , we will haveLocal minimum Analyzing the left and right of critical number x = 1 Since f ' ( x ) > 0 to left of x = 1 and f ' ( x ) < 0 to right of x = 1 ; At x = 1 , we will have a Local maximum Calculation (Continued): Description Steps Substituting x = − 1 in f ( x ) = x x 2 + 1 f ( − 1 ) = − 1 / ( ( − 1 ) 2 + 1 ) = − 1 / 2 = − 0.5 The local minimum point is ( − 1 , − 0.5 ) Substituting x = 1 in f ( x ) = x x 2 + 1 f ( 1 ) = 1 / ( 1 2 + 1 ) = 1 / 2 = 0.5 The local maximum point is ( 1 , 0

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Chapter 3, Problem 11SC

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Chapter 3, Problem 10SC

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Chapter 3 Solutions

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the roots, local extrema and inflection points of given function.

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Chapter 3 Solutions