Explanation Given: The function is f ( x ) = 1 x 2 − 9 . Definition used: (1) If the function gets larger and larger in magnitude without bound as x approaches the number k, then the line x = k is the vertical asymptote. (2) A function f ( x ) said to be symmetric about the y -axis if f ( − x ) = f ( x ) . Calculation: The given function is f ( x ) = 1 x 2 − 9 . Domain: Consider the given function f ( x ) = 1 x 2 − 9 . Simplify the function as follows, f ( x ) = 1 x 2 − 9 = 1 ( x + 3 ) ( x − 3 ) The function f ( x ) = 1 x 2 − 9 is defined at all the real number except the points x = − 3 and x = 3 . Thus, the domain of the function f ( x ) = 1 x 2 − 9 is ( − ∞ , − 3 ) ∪ ( − 3 , 3 ) ∪ ( 3 , ∞ ) . Symmetry: The given function is f ( x ) = 1 x 2 − 9 . Substitute x = − x in f ( x ) , f ( − x ) = 1 ( − x ) 2 − 9 = 1 x 2 − 9 Thus, f ( − x ) = f ( x ) . Therefore, from the definition stated above, the function does not have symmetry about the y -axis. Intercept: The function f ( x ) intercepts y -axis at x = 0 . Substitute the value x = 0 in f ( x ) , f ( 0 ) = 1 ( 0 ) 2 − 9 = − 1 9 That is the function f ( x ) intercepts y -axis at the point ( 0 , − 1 9 ) . The value of the function f ( x ) = 1 x 2 − 9 is never zero. Thus, the function f ( x ) does not intercept x -axis at any point. Asymptote: The function f ( x ) = 1 x 2 − 9 is not defined at the points x = − 3 and x = 3 . Thus, the function f ( x ) = 1 x 2 − 9 has vertical asymptotes at x = − 3 and x = 3 . The value of the function f ( x ) = 1 x 2 − 9 is never zero. Thus, the function f ( x ) has horizontal asymptotes at y = 0 . Critical point: The given function is f ( x ) = 1 x 2 − 9 . Differentiate f ( x ) with respect to x , f ′ ( x ) = − 2 x ( x 2 − 9 ) 2 For critical points, f ′ ( x ) = 0 . − 2 x ( x 2 − 9 ) 2 = 0 − 2 x = 0 x = 0 Thus, the critical point for the function f ( x ) = 1 x 2 − 9 is x = 0 . Substitute x = 0 in the function f ( x ) = 1 x 2 − 9 to obtain the critical point as ( 0 , − 1 9 ) . Relative Maxima and Relative Minima : The critical point for the function f ( x ) = 1 x 2 − 9 is at x = 0 . Substitute x = 4 in f ′ ( x ) = − 2 x ( x 2 − 9 ) 2 , f ′ ( 4 ) = − 2 × ( 4 ) ( ( 4 ) 2 − 9 ) 2 = − 8 ( 7 ) 2 = − 8 49 < 0 Thus, the function f ( x ) is decreasing on the interval ( 3 , ∞ ) . Substitute x = − 4 in f ′ ( x ) = − 2 x ( x 2 − 9 ) 2 , f ′ ( − 4 ) = − 2 × ( − 4 ) ( ( − 4 ) 2 − 9 ) 2 = 8 ( 7 ) 2 = 8 49 > 0 Thus, the function f ( x ) is increasing on the interval ( − ∞ , − 3 ) . Substitute x = 1 in f ′ ( x ) = − 2 x ( x 2 − 9 ) 2 , f ′ ( 1 ) = − 2 × ( 1 ) ( ( 1 ) 2 − 9 ) 2 = − 2 ( − 8 ) 2 = − 1 32 < 0 Thus, the function f ( x ) is decreasing on the interval ( 0 , 3 )

Calculus with Applications Books a la Carte Edition

11th Edition

ISBN: 9780133864564

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

Publisher: Pearson Education

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Calculus with Applications Books a la Carte Edition

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