Business Calc Homework w answers_Part_40

Business Calc Homework w answers_Part_40 - 196 Chapter 4...

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30. f ( x ) 5 e x 1 sin x f 9 ( x ) 5 e x 1 cos x L ( x ) 5 f (0) 1 f 9 (0)( x 2 0) 5 1 1 2( x 2 0) 5 2 x 1 1 31. The global minimum value of } 1 2 } occurs at x 5 2. 32. (a) The values of y 9 and y 0 are both negative where the graph is decreasing and concave down, at T . (b) The value of y 9 is negative and the value of y 0 is posi- tive where the graph is decreasing and concave up, at P . 33. (a) The function is increasing on the interval (0, 2]. (b) The function is decreasing on the interval [ 2 3, 0). (c) The local extreme values occur only at the endpoints of the domain. A local maximum value of 1 occurs at x 52 3, and a local maximum value of 3 occurs at x 5 2. 34. The 24th day 35. 36. (a) We know that f is decreasing on [0, 1] and increasing on [1, 3], the absolute minimum value occurs at x 5 1 and the absolute maximum value occurs at an endpoint. Since f (0) 5 0, f (1) 2, and f (3) 5 3, the absolute minimum value is 2 2 at x 5 1 and the absolute maxi- mum value is 3 at x 5 3. (b) The concavity of the graph does not change. There are no points of inflection. (c) 37. (a) f ( x ) is continuous on [0.5, 3] and differentiable on (0.5, 3). (b) f 9 ( x ) 5 ( x ) 1 } 1 x } 2 1 (ln x )(1) 5 1 1 ln x Using a 5 0.5 and b 5 3, we solve as follows. f 9 ( c ) 5 } f (3 3 ) 2 2 f 0 ( .5 0.5) } 1 1 ln c 5 } 3ln3 2 2. 0 5 .5 ln 0.5 } ln c 1 ln c 5 0.4 ln(27 ˇ 2 w ) 2 1 c 5 e 2 1 (27 ˇ 2 w ) 0.4 c 5 e 2 1 ˇ 5 1 w 4 w 5 w 8 w < 1.579 (c) The slope of the line is m 5 } f ( b b ) 2 2 f a ( a ) }5 0.4 ln (27 ˇ 2 w ) 5 0.2 ln 1458, and the line passes through (3, 3 ln 3). Its equation is y 5 0.2(ln 1458)( x 2 3) 1 3 ln 3, or approximately y 5 1.457 x 2 1.075. (d) The slope of the line is m 5 0.2 ln 1458, and the line passes through ( c , f ( c )) 5 ( e 2 1 ˇ 5 1 w 4 w 5 w 8 w , e 2 1 ˇ 5 1 w 4 w 5 w 8 w ( 2 1 1 0.2 ln 1458)) < (1.579, 0.722). Its equation is y 5 0.2(ln 1458)( x 2 c ) 1 f ( c ), y 5 0.2 ln 1458( x 2 e 2 1 ˇ 5 1 w 4 w 5 w 8 w ) 1 e 2 1 ˇ 5 1 w 4 w 5 w 8 w ( 2 1 1 0.2 ln 1458), y 5 0.2(ln 1458) x 2 e 2 1 ˇ 5 1 w 4 w 5 w 8 w , or approximately y 5 1.457 x 2 1.579. 38. (a) v ( t ) 5 s 9 ( t ) 5 4 2 6 t 2 3 t 2 (b) a ( t ) 5 v 9 ( t ) 6 2 6 t (c) The particle starts at position 3 moving in the positive direction, but decelerating. At approximately t 5 0.528, it reaches position 4.128 and changes direction, begin- ning to move in the negative direction. After that, it continues to accelerate while moving in the negative direction. 39. (a) L ( x ) 5 f (0) 1 f 9 (0)( x 2 0) 1 1 0( x 2 0) 1 (b) f (0.1) < L (0.1) 1 (c) Greater than the approximation in (b), since f 9 ( x ) is actually positive over the interval (0, 0.1) and the estimate is based on the derivative being 0. 40. (a) Since } d d y x } 5 ( x 2 )( 2 e 2 x ) 1 ( e 2 x )(2 x ) 5 (2 x 2 x 2 ) e 2 x , dy 5 (2 x 2 x 2 ) e 2 x dx .
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Business Calc Homework w answers_Part_40 - 196 Chapter 4...

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