Explanation Given: The function is f ( x ) = ln x 2 x 2 . Definition used: A function f ( x ) said to be symmetric about the y -axis if f ( − x ) = f ( x ) Calculation: The given function is f ( x ) = ln x 2 x 2 . Domain: The given function f ( x ) = ln x 2 x 2 is defined only for all real values except the point x = 0 Thus, the domain of the function f ( x ) = ln x 2 x 2 is ( − ∞ , 0 ) ∪ ( 0 , ∞ ) . Symmetry: The given function is f ( x ) = ln x 2 x 2 . Substitute x = − x in the function f ( x ) = ln x 2 x 2 , f ( − x ) = ln ( − x ) 2 ( − x ) 2 = ln x 2 x 2 Hence, f ( − x ) = f ( x ) . Therefore, from the definition stated above and conclude that the function is symmetric about the y -axis. Intercept: The function f ( x ) = ln x 2 x 2 is not defined at x = 0 . That is the function f ( x ) does not have any y - intercept. The function f ( x ) intercepts x -axis at f ( x ) = 0 . Substitute the value f ( x ) = 0 in f ( x ) , ln x 2 x 2 = 0 ln x 2 = 0 x 2 = e 0 x = ± 1 Thus, x = 1 and x = − 1 . That is the function f ( x ) intercepts x -axis at the point ( 1 , 0 ) and ( − 1 , 0 ) . Asymptote: The function f ( x ) = ln x 2 x 2 is not defined at the point x = 0 . Thus, the vertical asymptote of the function f ( x ) = ln x 2 x 2 is x = 0 . Thus, the function f ( x ) = ln x 2 x 2 has horizontal asymptote at y = 0 . Critical point: The given function is f ( x ) = ln x 2 x 2 . Differentiate f ( x ) with respect to x , f ′ ( x ) = x 2 × 2 x ( 1 x 2 ) − ln x 2 × ( 2 x ) ( x 2 ) 2 = 2 x ( 1 − ln x 2 ) x 4 = 2 ( 1 − ln x 2 ) x 3 To obtain critical points, f ′ ( x ) = 0 , 2 ( 1 − ln x 2 ) x 3 = 0 2 ( 1 − ln x 2 ) = 0 ln x 2 = 1 x 2 = e 1 Thus, x = − e and x = e . Thus, substitute x = − e and x = e in the function f ( x ) = ln x 2 x 2 to obtain the critical points as ( − e , 1 e ) and ( e , 1 e ) . Relative Maxima and Relative Minima : The critical points are at x = − e and x = e . Substitute x = 1 in f ′ ( x ) = 2 ( 1 − ln x 2 ) x 3 , f ′ ( 1 ) = 2 ( 1 − ln ( 1 ) 2 ) ( 1 ) 3 = 2 − 0 1 = 2 > 0 Thus, the function f ( x ) is increasing on the interval ( 0 , e ) . Substitute x = − 1 in f ′ ( x ) = 2 ( 1 − ln x 2 ) x 3 , f ′ ( − 1 ) = 2 ( 1 − ln ( − 1 ) 2 ) ( − 1 ) 3 = 2 − 0 − 1 = − 2 < 0 Thus, the function f ( x ) is decreasing on the interval ( − e , 0 ) . Substitute x = 4 in f ′ ( x ) = 2 ( 1 − ln x 2 ) x 3 , f ′ ( 4 ) = 2 ( 1 − ln ( 4 ) 2 ) ( 4 ) 3 = 2 ( 1 − l

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ISBN: 8220102020252

Author: RITCHEY

Publisher: PEARSON

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 13.4, Problem 24E

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EBK FINITE MATHEMATICS AND CALCULUS WIT

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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Solution Summary: The author explains that the function f(x) is symmetric about the y -axis.

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