Explanation Given data: y The given functions are: (i) g ( x ) = 6 x 2 − 18 x , (ii) g ( x ) = 2 − x − 2 x 2 + x 3 (iii) g ( x ) = ( x + 2 ) ( x − 1 ) 4 e x (iv) g ( x ) = 3 x 2 − 5 x − 2 sin 2 x Method/Formula used: The function is positive for the values of “ x ” for which g ( x ) > 0 , and negative for which g ( x ) < 0 . Calculation: (i).The function g ( x ) = 6 x 2 − 18 x , is positive at x when g ( x ) > 0 ⇒ 6 x 2 − 18 x > 0 ⇒ 6 x ( x − 3 ) > 0 The function g ( x ) = 6 x 2 − 18 x , is negative at x when g ( x ) < 0 ⇒ 6 x ( x − 3 ) < 0 For partition of real line into intervals, we put g ( x ) = 0 ⇒ 6 x ( x − 3 ) = 0 ⇒ x = 0 , x = 3 . Now, we draw sign chart to find the intervals in which function is positive or negative. Sign Chart of Function g ( x ) = 6 x 2 - 1 8 x Interval x ( x -3) g ( x ) x > 3 + + + 0 + - - x < 0 - - + From the sign chart of g ( x ) it is clear that the function is positive in intervals ( − ∞ , 0 ) and ( 3 , ∞ ) , and negative in the interval ( 0 , 3 ) . (ii).The function g ( x ) = 2 − x − 2 x 2 + x 3 , is positive at x when g ( x ) > 0 ⇒ 2 − x − 2 x 2 + x 3 > 0 ⇒ ( x − 1 ) ( x + 1 ) ( x − 2 ) > 0 The function g ( x ) = 2 − x − 2 x 2 + x 3 , is negative at x when g ( x ) < 0 or ( x − 1 ) ( x + 1 ) ( x − 2 ) < 0 For partition of real line into intervals, we put g ( x ) = 0 ⇒ ( x − 1 ) ( x + 1 ) ( x − 2 ) = 0 ⇒ x = − 1 , x = 1 , x = 2 . Now, we draw sign chart to find the intervals in which function is positive or negative. Sign Chart of Function g ( x ) = 2 - x - 2 x 2 + x 3 Interval ( x -2) ( x -1) ( x +1) g ( x ) x < -1 - - - - -1 - - + + 1 - + + - x> 2 + + + + From the sign chart of g ( x ) it is clear that the function is negative in intervals ( − ∞ , − 1 ) and ( 1 , 2 ) , and positive in the intervals ( − 1 , 1 ) and ( 2 , ∞ ) . (iii).The function g ( x ) = ( x + 2 ) ( x − 1 ) 4 e x , is positive at x when g ( x ) > 0 . Since, the functions e x and ( x − 1 ) 4 are always on the real line, therefore, g ( x ) > 0 ⇒ ( x + 2 ) > 0 ⇒ x > − 2

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 3.2, Problem 3TB

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The interval in which function is positive and in which negative, using algebra and sign chart.

Solution Summary: The author explains the function g(x)=6x218x, which is positive in intervals (,0) and (3,).

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