Explanation Given information: The function is g ( x ) = 1 x − x 4 4 . Concept use: The domain of a function is the set of all possible values of x such that the function exists at those values. Calculation: The given function is g ( x ) = 1 x − x 4 4 . Factorize the above function as, g ( x ) = 1 x − x 4 4 = 1 x ( 1 − x 3 ) 4 = 1 x ( 1 − x ) ( 1 + x + x 2 ) 4 Consider the polynomial x ( 1 − x ) ( 1 + x + x 2 ) as r ( x ) , then the function can be written as, f ( x ) = 1 r ( x ) 4 The polynomial can be written as, r ( x ) = x ( 1 − x ) ( 1 + x + x 2 ) Here, f ( x ) is defined only for positive values of the polynomial x ( 1 − x ) ( 1 + x + x 2 ) . Therefore, the inequality becomes r ( x ) > 0 . The zero of the polynomial r ( x ) are calculated as, r ( x ) = 0 x ( 1 − x ) ( 1 + x + x 2 ) = 0 x = 0 or 1- x = 0 or 1+ x + x 2 = 0 x = 0 or x = 1 or 1+ x + x 2 = 0 Therefore, the real zeros of the polynomial r ( x ) are 0 and 1 , as the polynomial 1+ x + x 2 has no real zeros. Hence, the intervals determined by the zeros of the polynomial are ( − ∞ , 0 ) , ( 0 , 1 ) and ( 1 , ∞ ) . The sign of each factor, in each interval are shown in Table 1. Sign of ( − ∞ , 0 ) ( 0 , 1 ) ( 1 , ∞ ) x − + + 1 − x + + − 1 + x + x 2 + + + r ( x ) = x ( 1 − x ) ( 1 + x + x 2 ) − + − Table 1 From the Table 1, it is shown that the inequality r ( x ) > 0 satisfies in the interval ( 0 , 1 ) . The end points of the intervals ( − ∞ , 0 ) , ( 0 , 1 ) and ( 1 , ∞ ) , are 0 and 1 . To check whether the end points satisfies the inequality r ( x ) > 0 or not, substitute the end points for x in the inequality r ( x ) > 0 respectively

7th Edition

ISBN: 9781305115545

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 3, Problem 102E

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The domain of the function g(x)=1x−x44.

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Chapter 3, Problem 101E

Chapter 3, Problem 103E

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