Explanation Given information: The functions are, y = x 2 3 y = x 4 3 Graph: To graph of the function y = x 2 3 , first obtain the x -intercept of the function. The function y = x 2 3 is defined for all real values of x such that the domain of the function is ( − ∞ , ∞ ) . To obtain the x -intercept of the function, substitute 0 for y in the function as, x 2 3 = 0 x = 0 Therefore, the x -intercept of the function y = x 2 3 is ( 0 , 0 ) . The domain of the function is ( − ∞ , ∞ ) and the function intersects the x -axis at only one point ( 0 , 0 ) such that the defined interval is ( − ∞ , 0 ) and ( 0 , ∞ ) . To check that the function has symmetry about y -axis, replace − x by x in the function y = x 2 3 as, y = ( − x ) 2 3 y = x 2 3 Therefore, the function y = x 2 3 has symmetry about y -axis. Now, differentiate the function y = x 2 3 with respect to x as, d y d x = d d x x 2 3 = 2 3 x 2 3 − 1 = 2 3 x − 1 3 = 2 3 x 1 3 The slope of the function y = x 2 3 is not defined for x = 0 . Consider a test point x = − 27 on the interval ( − ∞ , 0 ) to check that the derivative of the function is positive or negative. Now, substitute − 27 for x in the derivative of the function y = x 2 3 to check the sign of the derivative of the function. d y d x = 2 3 ( − 27 ) 1 3 = 2 3 ( − 3 ) = − 2 9 The derivative of the function y = x 2 3 is negative on the interval ( − ∞ , 0 ) therefore, the slope is decreasing on the interval ( − ∞ , 0 ) . Consider a test point x = 27 on the interval ( 0 , ∞ ) to check that the derivative of the function is positive or negative. Now, substitute 27 for x in the derivative of the function y = x 2 3 to check the sign of the derivative of the function. d y d x = 2 3 ( 27 ) 1 3 = 2 3 ( 3 ) = 2 9 The derivative of the function y = x 2 3 is positive on the interval ( 0 , ∞ ) therefore, the slope is increasing on the interval ( 0 , ∞ ) . To check the concavity of the function y = x 2 3 , differentiate the derivative of the function y = x 2 3 with respect to x as, d y d x ( d y d x ) = d d x ( 2 3 x 1 3 ) d 2 y d x 2 = 2 3 ( − 1 3 ) x − 1 3 − 1 d 2 y d x 2 = − 2 9 x − 4 3 d 2 y d x 2 = − 2 9 x 4 3 The second derivative of the function y = x 2 3 is − 2 9 x 4 3 . It is clearly seen that the denominator 9 x 4 3 of the second derivative of function y = x 2 3 is always positive for any value of x such that the second derivative of the function is always negative. If the second derivative of the function y = x 2 3 is negative on the interval ( − ∞ , 0 ) and ( 0 , ∞ ) such that the shape of the function is concave down in the interval ( − ∞ , 0 ) and ( 0 , ∞ ) . Thus, the graph of the function y = x 2 3 is shown in Figure 1

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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