Explanation Given: The function is, f ( x ) = x 2 e 2 x . Definition used: A function f ( x ) is said to be symmetric about the y -axis if f ( − x ) = f ( x ) . Calculation: Domain: The given function f ( x ) = x 2 e 2 x is defined for all real values. Thus, the domain of the function f ( x ) = x 2 e 2 x is ( − ∞ , ∞ ) . Symmetry: The given function is f ( x ) = x 2 e 2 x . Substitute x = − x in f ( x ) , f ( − x ) = ( − x ) 2 e 2 ( − x ) = x 2 e − 2 x Thus, f ( − x ) ≠ f ( x ) . Therefore, from the definition stated above and conclude that the polynomial has no symmetry. Intercept: The function f ( x ) intercepts y -axis at x = 0 . Substitute the value x = 0 in f ( x ) , f ( 0 ) = ( 0 ) 2 e 2 ( 0 ) = 0 That is the function f ( x ) intercepts y -axis at the point ( 0 , 0 ) . The function f ( x ) intercepts x -axis at f ( x ) = 0 . Substitute the value f ( x ) = 0 in f ( x ) , x 2 e 2 x = 0 The function e 2 x is never 0. Thus, x = 0 . That is the function f ( x ) intercepts x -axis at the point ( 0 , 0 ) . Asymptote: The given function is f ( x ) = x 2 e 2 x . Calculate the limit of the function, lim x → ∞ f ( x ) = lim x → ∞ x 2 e 2 x = 0 ( ∵ lim x → ∞ x n e n x = 0 ) Thus, the function f ( x ) = x 2 e 2 x has horizontal asymptote at y = 0 . Critical point: The given function is f ( x ) = x 2 e 2 x . Differentiate f ( x ) with respect to x , f ′ ( x ) = ( 2 x ) ( e 2 x ) + ( x 2 ) ( 2 e 2 x ) = 2 ( x 2 + x ) e 2 x = 2 x ( x + 1 ) e 2 x To obtain critical points, f ′ ( x ) = 0 , 2 x ( x + 1 ) e 2 x = 0 x ( x + 1 ) = 0 x = 0 or x = − 1 Thus, x = 0 or x = − 1 . Thus, the critical points are at x = 0 and x = − 1 . Substitute x = 0 and x = − 1 in the function f ( x ) = x 2 e 2 x to obtain the critical points as ( − 1 , e − 2 ) and ( 0 , 0 ) . Relative Maxima and Relative Minima : The critical points are at x = 0 and x = − 1 . Substitute x = − 2 in f ′ ( x ) = 2 x ( x + 1 ) e 2 x , f ′ ( − 2 ) = 2 ( − 2 ) ( − 2 + 1 ) e 2 ( − 2 ) = 4 e − 4 > 0 Thus, the function f ( x ) is increasing on the interval ( − ∞ , − 1 ) . Substitute x = − 1 2 in f ′ ( x ) = 2 x ( x + 1 ) e 2 x , f ′ ( − 1 2 ) = 2 ( − 1 2 ) ( − 1 2 + 1 ) e 2 ( − 1 2 ) = − 1 2 e − 1 < 0 Thus, the function f ( x ) is decreasing on the interval ( − 1 , 0 )

Calculus with Applications (11th Edition)

11th Edition

ISBN: 9780321979421

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 5, Problem 54RE

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To sketch: The graph of the function f(x)=x2e2x .

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Chapter 5, Problem 53RE

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Chapter 5 Solutions

Calculus with Applications (11th Edition)

Ch. 5.1 - YOUR TURN 1 Find where the function is increasing...Ch. 5.1 - Prob. 2YT Ch. 5.1 - Prob. 3YT Ch. 5.1 - Prob. 4YT Ch. 5.1 - Prob. 1WE Ch. 5.1 - Prob. 2WE Ch. 5.1 - Prob. 3WE Ch. 5.1 - Prob. 4WE Ch. 5.1 - Find the derivative of each of the following...Ch. 5.1 - Prob. 6WE

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The graph of the function f ( x ) = x 2 e 2 x .

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To sketch: The graph of the function f(x)=x2e2x .

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