Explanation Given information : Critical point and Local extremum Concept Involved: The point ( c , f ( c ) ) is called a critical point of a function f ( x ) if c is in the domain of function and either it makes first derivative a zero { f ' ( c ) = 0 } or undefined { f ' ( c ) = D o e s N o t E x i s t }. To find the critical numbers of the function, we need to identify the point from the graph where the f ' graph cuts the x -axis. Once we find the critical numbers as per above definition, we need to list possible intervals where the function is increasing or decreasing using the critical numbers. In the f ' graph, the interval where f ' < 0 {below the x -axis} is the interval where the function decreasing. In the f ' graph, the interval where f ' > 0 {above the x -axis} is the interval where the function increasing. 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 ) ) Formula Used: Exponent rule: a m n = a m / n Product rule: ( u ⋅ v ) ' = u ⋅ v ' + v ⋅ u ' Power rule of derivative: If f ( x ) = k x n then f ' ( x ) = k ⋅ n x n − 1 Chain rule: ( f ( g ( x ) ) ) ' = f ' ( g ( x ) ) ⋅ g ' ( x ) Example: Consider a function f ( x ) = x 4 − x 2 Calculation: Description Steps Rewrite the given function with rational exponent f ( x ) = x ( 4 − x 2 ) 1 / 2 Domain of the function will be the values of x that makes the expression inside the square root greater than or equal to zero. 4 − x 2 ≥ 0 Factor the left side of the equation ( 2 − x ) ( 2 + x ) ≥ 0 Set each factor to zero and solve for xto find the boundary points x = − 2 , 2 Write the possible solution interval ( − ∞ , − 2 ] , [ − 2 , 2 ] , [ 2 , ∞ ) Pick a test point from each interval and find test point from which intervals make the inequality true. x = − 3 , will make inequality False. x = 0 , will make inequality True. x = 3 , will make inequality False. So, [ − 2 , 2 ] is the domain of the function Use the product rule, chain rule and power rule to find the derivative of the function x ⋅ d ( ( 4 − x 2 ) 1 / 2 ) d x + ( 4 − x 2 ) 1 / 2 ⋅ d ( x ) d x f ' ( x ) = x ⋅ 1 2 ( 4 − x 2 ) 1 2 − 1 ⋅ d ( 4 − x 2 ) d x + ( 4 − x 2 ) 1 / 2 ⋅ 1 = x 2 ( 4 − x 2 ) − 1 2 ⋅ ( − 2 x ) + ( 4 − x 2 ) 1 / 2 ⋅ 1 = 4 − x 2 1 − x 2 4 − x 2 = 4 − x 2 4 − x 2 − x 2 4 − x 2 = 4 − 2 x 2 4 − x 2 f ' ( x ) will become zero if we set numerator to zero 4 − 2 x 2 = 0 Subtract 4 on both sides − 2 x 2 = − 4 Divide by -2 on both sides x 2 = 2 Take square root on both sides x = ± 2 f ' ( x ) will become undefined if we set denominator to zero 4 − x 2 = 0 Square on both sides of the equation 4 − x 2 = 0 Subtract 4 on both sides − x 2 = − 4 Divide by -1 on both sides x 2 = 4 Description Steps Taking square root on both sides of equation x 2 = 4 x = ± 2 List all the critical numbers of the function that makes derivative of function zero or undefined x = − 2 , − 2 , 2 , & 2 List all the intervals where the function will be increasing or decreasing using the critical number

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Confidence Intervals

When we calculate intervals in probability and statistics, it can be simply termed as finding out a confidence interval. The confidence interval is usually calculated from the data set which is observed. The value of the parameter that needs to be calcul…

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critical point intuitively and describe what is local extremum.

Solution Summary: The author explains the concept of critical point and local extremum by identifying the point where the graph cuts the x -axis.

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Chapter 3 Solutions