Explanation Given information: The provided function is: f ( x ) = x e − x 2 Formula used: Stationary Points: To locate stationary points put f ′ ( x ) = 0 and solve for x . Second derivative test: If the function f ( x ) has a stationary point at x = c and that f ″ ( c ) exists then evaluate the sign of f ″ ( c ) : A. If f ″ ( c ) < 0 , then f has a relative maximum at x = c . B. If f ″ ( c ) > 0 , then f has a relative minimum at x = c . Also, if f ″ ( c ) = 0 then the test is fails so use the first derivative test. Product rule of differentiation: d d x ( a . b ) = a ' b + a b ' Where a and b are functions. Calculation: The provided function is: f ( x ) = x e − x 2 Differentiate the above equation with respect to x to get: f ′ ( x ) = d d x ( x e − x 2 ) Consider the product rule of differentiation: d d x ( a . b ) = a ′ b + a b ′ Substitute x for a and e − x 2 for b in above equation to get: d d x ( a . b ) = ( x ) ' ( e − x 2 ) + ( x ) ( e − x 2 ) ' = e − x 2 + x ( − 2 x ) e − x 2 = e − x 2 [ 1 − 2 x 2 ] Consider the formula for stationary points: f ′ ( x ) = 0 Substitute e − x 2 [ 1 − 2 x 2 ] for f ' ( x ) to get: e − x 2 [ 1 − 2 x 2 ] = 0 As e − x 2 cannot be zero, Thus, 1 − 2 x 2 = 0 2 x 2 = 1 x 2 = 1 2 Thus, x = 1 2 And, x = − 1 2 Since, f ′ ( x ) = 0 at x = 1 2 , − 1 2 . Thus, the x − coordinate of the critical point is x = 1 2 , − 1 2 . Differentiate f ′ ( x ) with respect to x to get: f ″ ( x ) = d d x ( e − x 2 [ 1 − 2 x 2 ] ) Consider the product rule of differentiation: d d x ( a

Bundle: Applied Calculus, Loose-leaf Version, 7th + Webassign Printed Access Card For Waner/costenoble's Applied Calculus, 7th Edition, Single-term

7th Edition

ISBN: 9781337604703

Author: Waner, Stefan; Costenoble, Steven

Publisher: Cengage Learning

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Chapter 5.3, Problem 44E

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To calculate: The x− coordinates of all the critical points of the function f(x)=xe−x2 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

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Chapter 5.3, Problem 43E

Chapter 5.3, Problem 45E

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The x − coordinates of all the critical points of the function f ( x ) = x e − x 2 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

Solution Summary: The author explains how to calculate the x- coordinates of all the critical points of the function.

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To calculate: The x− coordinates of all the critical points of the function f(x)=xe−x2 and check by first using the second derivative test and by some other method if the test fails, whether the each critical point is relative minimum, relative maximum or neither.

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Chapter 5 Solutions