Explanation Definition used: If the derivative of function f ' ( x ) changes sign from positive to negative as the function’s graph move across a critical number c , then function f has relative maximum at x = c . If the derivative of function f ' ( x ) changes sign from negative to positive as the function’s graph move across a critical number c , then function f has relative minimum at x = c . Given information: The given function is f ( x ) = 1 2 x 4 − x 2 . Calculation: To get the relative minimum and maximum values first find the derivative of function. d d x f ( x ) = d d x ( 1 2 x 4 − x 2 ) = d d x ( 1 2 x 4 ) − d d x ( x 2 ) = 2 x 3 − 2 x = 2 x ( x + 1 ) ( x − 1 ) [ ∵ a 2 − b 2 = ( a + b ) ( a − b ) ] The derivative of function is f ′ ( x ) = 2 x ( x + 1 ) ( x − 1 ) . To get the zero, set the derivative of function is equal to zero. f ′ ( x ) = 0 2 x ( x + 1 ) ( x − 1 ) = 0 The zeroes of the derivative of function are x = − 1 , x = 0 and x = 1 . The critical points of the function are x = − 1 , x = 0 and x = 1 . To get the intervals of function, find the domain of function f ( x ) = 1 2 x 4 − x 2 . The function is not undefined at any point, thus, the domain of function is ( − ∞ , ∞ ) . Combine the critical points x = − 1 , x = 0 and x = 1 with the domain. The intervals of the function are ( − ∞ , − 1 ) , ( − 1 , 0 ) , ( 0 , 1 ) and ( 1 , ∞ ) . Select a point in each interval, to check the sign of the derivative of function. Interval Test point f ′ ( x ) = 2 x ( x + 1 ) ( x − 1 ) Sign of f ′ ( x ) ( − ∞ , − 1 ) − 2 − 12 − ( − 1 , 0 ) − 1 2 3 4 + ( 0 , 1 ) 1 2 − 3 4 − ( 1 , ∞ ) 2 12 + Function f ′ ( x ) changes sign from negative to positive as the function move across the critical point x = − 1 from left to right that means the function has relative minimum at x = − 1

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Applied Calculus for the Managerial, Life, and Social Sciences

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