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# hw8 - Math3C HW#8 [email protected]

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Unformatted text preview: Math3C HW#8 [email protected] [email protected] Exercise 1 Ex11 Solution: ( a ) We need to show: f ( u & x ) = f ( u + x ) consider f ( x ) = 1 & p 2 Â¡ e & ( x & u ) 2 = ( 2 & 2 ) f ( u & x ) = f ( u + x ) = 1 & p 2 Â¡ e & x 2 = ( 2 & 2 ) ( b ) since f ( x ) Â¡ 1 & p 2 Â¡ e ; and f ( u ) = 1 & p 2 Â¡ e we have for all x, f ( x ) Â¡ f ( u ) : ( c ) f 00 ( x ) = d dx 2 & 1 & p 2 Â¡ e & ( x & u ) 2 = ( 2 & 2 ) Â¡ = & 1 2 p 2 p Â¡& 5 exp & & 1 2 & 2 ( u & x ) 2 Â¡ Â¢ & u 2 + 2 ux & x 2 + & 2 Â£ = f ( x ) & 4 & ( x & u ) 2 & & 2 Â¡ = 0 ) u 2 + 2 ux & x 2 = & 2 x & u = Â¢ & ) x = u Â¢ &: ( d ) The graph of f ( x ) = 1 p 2 Â¡ e & ( x & 2) 2 = 2 is:-2-1 1 2 3 4 5 6 0.1 0.2 0.3 0.4 x y & Exercise 2 Ex12 1 Solution: We need to show: u = R 1 &1 xf ( x ) dx R 1 &1 x ( fx ) dx = R 1 &1 x 1 & p 2 Â¡ e & ( x & u ) 2 = ( 2 & 2 ) dx Let y = x & u & Use substitution rule: ) R 1 &1 x ( fx ) dx = R 1 &1 ( &y + u ) e & y 2 2 1 p 2 Â¡& &dy = ( &y + u ) e & y 2 2 1 p 2 Â¡& = R 1 &1 e & y 2 2 &y 1 p...
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