Explanation Given: The function is f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 . Definition used: A function f ( x ) said to be symmetric about the y -axis if f ( − x ) = f ( x ) . Calculation: Domain: The given function f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 is defined for all real values. Thus, the domain of the function f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 is ( − ∞ , ∞ ) . Symmetry: The given function is f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 . Substitute x = − x in f ( x ) , f ( − x ) = ( − x ) 4 − 4 3 ( − x ) 3 − 4 ( − x ) 2 + 1 = x 4 + 4 3 x 3 − 4 x 2 + 1 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 ) 4 − 4 3 ( 0 ) 3 − 4 ( 0 ) 2 + 1 = 1 That is the function f ( x ) intercepts y -axis at the point ( 0 , 1 ) . Asymptote: The function f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 has no asymptote. Critical point: The given function is f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 . Differentiate f ( x ) with respect to x , f ′ ( x ) = d d x ( x 4 − 4 3 x 3 − 4 x 2 + 1 ) = 4 x 3 − 4 3 ( 3 x 2 ) − 4 ( 2 x ) = 4 x 3 − 4 x 2 − 8 x To obtain critical points, f ′ ( x ) = 0 , 4 x 3 − 4 x 2 − 8 x = 0 4 x ( x 2 − x − 2 ) = 0 x ( x − 2 ) ( x + 1 ) = 0 Thus, x = 0 , x = 2 and x = − 1 . Thus, the critical pointsare at x = 0 , x = 2 and x = − 1 . Substitute x = 0 , x = 2 and x = − 1 in f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 and obtain the critical points as ( 0 , 1 ) , ( 2 , − 29 3 ) and ( − 1 , − 2 3 ) . Inflection Point: Differentiate the function f ′ ( x ) = 4 x 3 − 4 x 2 − 8 x f ′ ′ ( x ) = d d x ( 4 x 3 − 4 x 2 − 8 x ) = 4 ( 3 x 2 ) − 4 ( 2 x ) − 8 = 12 x 2 − 8 x − 8 To obtain inflection points, f ″ ( x ) = 0 , 12 x 2 − 8 x − 8 = 0 3 x 2 − 2 x − 2 = 0 x = 2 ± 4 − ( − 24 ) 6 x = 1 ± 7 3 Thus, the inflection pointsare x = 1 + 7 3 and x = 1 − 7 3 . Substitute x = 1 + 7 3 and x = 1 − 7 3 in f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 and obtain the inflection pointsas ( 1 + 7 3 , − 5.12 ) and ( 1 − 7 3 , 0.11 ) Increasing and decreasing: The function f ( x ) = x 4 − 4 3 x 3 − 4 x 2 + 1 has critical points x = 0 , x = 2 and x = − 1 . Substitute x = − 2 in f ′ ( x ) = 4 x 3 − 4 x 2 − 8 x , f ′ ( − 2 ) = 4 ( − 2 ) 3 − 4 ( − 2 ) 2 − 8 ( − 2 ) = − 24 − 16 + 16 = − 24 < 0 Thus, the function f ( x ) is decreasing on the interval ( − ∞ , − 1 )

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

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

Publisher: PEARSON

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Finite Mathematics and Calculus with Applications (10th Edition)

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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