Pre-Calc Homework Solutions 342

Pre-Calc Homework Solutions 342 - 50(a f x 5 ˇ 1 2 w p w e...

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Unformatted text preview: 50. (a) f ( x ) 5 } ˇ 1 2 w p w } e 2 x 2 /2 [ 2 3, 3] by [0, 0.5] f is increasing on ( 2‘ , 0]. f is decreasing on [0, ‘ ). f has a local maximum at (0, f (0)) 5 1 0, } ˇ 1 2 w p w } 2 (b) NINT 1 } ˇ 1 2 w p w } e 2 x 2 /2 , x , 2 1, 1 2 < 0.683 NINT 1 } ˇ 1 2 w p w } e 2 x 2 /2 , x , 2 2, 2 2 < 0.954 NINT 1 } ˇ 1 2 w p w } e 2 x 2 /2 , x , 2 3, 3 2 < 0.997 (c) Part (b) suggests that as b increases, the integral approaches 1. We can make E b 2 b f ( x ) dx as close to 1 as we want by choosing b . 1 large enough. Also, we can make E ‘ b f ( x ) dx and E 2 b 2‘ f ( x ) dx as small as we want by choosing b large enough. This is because 0 , f ( x ) , e 2 x /2 for x . 1. (Likewise, 0 , f ( x ) , e x /2 for x , 2 1) Thus, E ‘ b f ( x ) dx , E ‘ b e 2 x /2 dx E ‘ b e 2 x /2 dx 5 lim c → ‘ E c b e 2 x /2 dx 5 lim c → ‘ 3 2 2 e 2 x /2 4 5 lim c → ‘ [ 2 2 e 2 c /2 1 2 e 2 b /2 ] 5 2 e 2 b /2 As b → ‘ , 2 e 2 b /2 →0, so for large enough b , E ‘ b f ( x ) dx...
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