Unformatted text preview: ‘’ ( x ) = ( e(( xa)^2)/2b )(2( xa )/(2 b )) + ( e(( xa)^2)/2b )(2/ b ) Which when set equal to zero gives us: (2/ b ) e(( xa)^2)/2b = (x / b + a / b ) ( e(( xa)^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....
View
Full Document
 Spring '08
 Muralee
 Calculus, Derivative, Mathematical analysis, Convex function, absolute max, JP Rodriguez Prof.

Click to edit the document details