Explanation Given: The function is f ( x ) = 1 x 2 + 4 . 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 origin if f ( − x ) = − f ( x ) . Calculation: The given function is f ( x ) = 1 x 2 + 4 . Domain: The given function f ( x ) = 1 x 2 + 4 is defined at all the real numbers. Thus, the domain of the function f ( x ) = 1 x 2 + 4 is ( − ∞ , ∞ ) . Symmetry: The given function is f ( x ) = 1 x 2 + 4 . Substitute x = − x in f ( x ) , f ( − x ) = 1 ( − x ) 2 + 4 = 1 x 2 + 4 Thus, f ( − x ) = − f ( x ) . Therefore, from the definition stated above, the polynomial is symmetric about the origin. Intercept: The function f ( x ) intercepts y -axis at x = 0 . Substitute the value x = 0 in f ( x ) , f ( 0 ) = 1 ( 0 ) 2 + 4 = 1 4 That is the function f ( x ) intercepts y -axis at the point ( 0 , 1 4 ) . The value of the function f ( x ) = 1 x 2 + 4 is never zero. Thus, the function f ( x ) = 1 x 2 + 4 does not intercept x -axis. Asymptote: The function f ( x ) = 1 x 2 + 4 does not have any vertical asymptote. Thus, as the value of x tends to infinity, the graph of the function f ( x ) = 1 x 2 + 4 tends to the line y = 0 . Thus y = 0 is the horizontal asymptote for the function f ( x ) = 1 x 2 + 4 . Critical point: The given function is f ( x ) = 1 x 2 + 4 . Differentiate f ( x ) with respect to x , f ′ ( x ) = − 2 x ( x 2 + 4 ) 2 To obtain critical points, f ′ ( x ) = 0 , − 2 x ( x 2 + 4 ) 2 = 0 2 x = 0 x = 0 Thus, x = 0 . Thus, the critical point is at x = 0 . Substitute x = 0 in f ( x ) = 1 x 2 + 4 and obtain the critical points as ( 0 , 1 4 ) . Relative Maxima and Relative Minima : The critical point is at x = 0 . Substitute x = 1 in f ′ ( x ) = − 2 x ( x 2 + 4 ) 2 , f ′ ( 1 ) = − 2 × ( 1 ) ( ( 1 ) 2 + 4 ) 2 = − 2 ( 5 ) 2 = − 2 25 < 0 Thus, the function f ( x ) is decreasing on the interval ( 0 , ∞ ) . Substitute x = − 1 in f ′ ( x ) = − 2 x ( x 2 + 4 ) 2 , f ′ ( − 1 ) = − 2 × ( − 1 ) ( ( − 1 ) 2 + 4 ) 2 = 2 ( 5 ) 2 = 2 25 > 0 Thus, the function f ( x ) is increasing on the interval ( − ∞ , 0 )

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 13.4, Problem 18E

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To sketch: The graph of the function.

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Chapter 13.4, Problem 17E

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Chapter 13 Solutions

Finite Mathematics and Calculus with Applications

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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The graph of the function.

Solution Summary: The function f(x) is symmetric about the origin if it gets larger in magnitude without bound as x approaches the number k.

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Chapter 13 Solutions