To determine To sketch: The graph of the function. Explanation Given: The function is f ( x ) = x 1 3 + x 4 3 . 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 ) = x 1 3 + x 4 3 . Domain: The given function f ( x ) = x 1 3 + x 4 3 is defined for all real values. Thus, the domain of the function f ( x ) = x 1 3 + x 4 3 is ( − ∞ , ∞ ) . Symmetry: The given function is f ( x ) = x 1 3 + x 4 3 . Substitute x = − x in f ( x ) , f ( − x ) = ( − x ) 1 3 + ( − x ) 4 3 = x 1 3 − x 4 3 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 ) 1 3 + ( 0 ) 4 3 = 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 1 3 + x 4 3 = 0 x 1 3 ( 1 + x ) = 0 Thus, x = 0 and x = − 1 . That is the function f ( x ) intercepts x -axis at the point ( 0 , 0 ) and ( − 1 , 0 ) . Asymptote: The function f ( x ) = x 1 3 + x 4 3 has no asymptote. Critical point: The given function is f ( x ) = x 1 3 + x 4 3 . Differentiate f ( x ) with respect to x , f ′ ( x ) = 1 3 x − 2 3 + 4 3 x 1 3 = 1 3 x − 2 3 ( 1 + 4 x ) = 1 + 4 x 3 x 2 3 To obtain critical points, f ′ ( x ) = 0 , 1 + 4 x 3 x 2 3 = 0 1 + 4 x = 0 x = − 1 4 Thus, x = − 1 4 . Thus, the critical point is at x = − 1 4 . Substitute x = − 1 4 in the function f ( x ) = x 1 3 + x 4 3 to obtain the critical point as ( − 1 4 , − 0.472 ) . Relative Maxima and Relative Minima : The critical point is at x = − 1 4 . But f ′ ( x ) = 1 + 4 x 3 x 2 3 is not defined at the point x = 0 . Substitute x = 1 in f ′ ( x ) = 1 + 4 x 3 x 2 3 , f ′ ( 1 ) = 1 + 4 × ( 1 ) 3 × ( 1 ) 2 3 = 5 3 > 0 Thus, the function f ( x ) is increasing on the interval ( 0 , ∞ ) . Substitute x = − 1 in f ′ ( x ) = 1 + 4 x 3 x 2 3 , f ′ ( − 1 ) = 1 + 4 × ( − 1 ) 3 × ( − 1 ) 2 3 = − 3 3 = − 1 < 0 Thus, the function f ( x ) is decreasing on the interval ( − ∞ , − 1 4 )

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 5.4, Problem 30E

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To sketch: The graph of the function.

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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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