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Unformatted text preview: ( x ) = ( e-(( x-a)^2)/2b )(2( x-a )/(2 b )) + ( e-(( x-a)^2)/2b )(-2/ b ) Which when set equal to zero gives us: (2/ b ) e-(( x-a)^2)/2b = (-x / b + a / b ) ( e-(( x-a)^2)/2b ) After cancelling out everything that can be cancelled, we are left with x = a 2 When we plug this value of x into the first derivative, we get the answer -2/b, meaning that the graph is concave down and there is an absolute max at x = 2 C) f ( x ) approaches zero as x goes to negative infinity. The line concaves upwards near the point x = -1.5. We find the absolute maximum at (2,1) , then the line concaves downwards again. f ( x ) again approaches zero, but this time as x goes to positive infinity....
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This note was uploaded on 04/13/2008 for the course MATH 021 taught by Professor Muralee during the Spring '08 term at Lehigh University .
- Spring '08