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Unformatted text preview: Section 4.1 13. f ( x ) = 3 . The derivative is positive for all x so that the function is increasing on ( , + ) and the function is never decreasing. 15. f ( x ) = 2 x 3 = 0 x = 3 / 2 is the only critical number at which f ( x ) = 0. There are no critical numbers at which f ( x ) is undefined. Evaluating f ( x ) at test points on either side of the critical number we find f (0) = 3 < 0 and f (2) = 1 > . Thus the function is decreasing on ( , 3 / 2) and increasing on (3 / 2 , + ) . 17. g ( x ) = 1 3 x 2 = 0 x = q 1 / 3 = ( 3 / 3) are the two critical numbers at which g ( x ) = 0. There are no critical numbers at which g ( x ) is undefined. Evaluating g ( x ) at test points on either side of each critical number we find g ( 1) = 2 < , g (0) = 1 > , and g (1) = 2 < . Thus the function is decreasing on ( , q 1 / 3) ( q 1 / 3 , + ) and increasing on ( q 1 / 3 , q 1 / 3) . 19. g ( x ) = 3 x 2 + 6 x = 3 x ( x + 2) = 0 x = 2 , 0 are the two critical numbers at which g ( x ) = 0. There are no critical numbers at which g ( x ) is undefined. Evaluating g ( x ) at test points on either side of each critical number we find g ( 3) = 9 > , g ( 1) = 3 < , and g (1) = 9 > . Thus the function is decreasing on ( 2 , 0) and increasing on ( , 2) (0 , + ) . 21. f ( x ) = x 2 6 x + 9 = ( x 3) 2 = 0 x = 3 is the only critical number at which f ( x ) = 0. There are no critical numbers at which f ( x ) is undefined. Evaluating f ( x ) at test points on either side of the critical number we find f (2) = 1 > 0 and f (4) = 1 > . Thus the function is increasing on ( , 3) (3 , + ) and the function is never decreasing. 23. h ( x ) = 4 x 3 12 x 2 = 4 x 2 ( x 3) = 0 x = 0 , 3 are the two critical numbers at which h ( x ) = 0. There are no critical numbers at which h ( x ) is undefined. Evaluating h ( x ) at test points on either side of each critical number we find h ( 1) = 16 < , h (1) = 8 < , and h (4) = 64 > . Thus the function is decreasing on ( , 0) (0 , 3) and increasing on (3 , + ) . 25. f ( x ) = 1 ( x 2) 2 . The function has no critical numbers at which the derivative is equal to zero, however, there is a critical number at x = 2 at which f ( x ) is undefined. Evaluating f ( x ) at test points on either side of the critical number we find f (1) = 1 < 0 and f (3) = 1 < . Thus the function is decreasing on ( , 2) (2 , + ) and the function is never increasing. 27. h ( t ) = 1 ( t 1) 2 . The function has no critical numbers at which the derivative is equal to zero, however, there is a critical number at t = 1 at which h ( t ) is undefined. Evaluating h ( t ) at test points on either side of the critical number we find h (0) = 1 < 0 and h (2) = 1 < . Thus the function is decreasing on ( , 1) (1 , + ) and the function is never increasing....
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 Spring '08
 Smith
 Calculus, Derivative

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