written assignment 5

written assignment 5 - ( x ) = ( e-(( x-a)^2)/2b )(2( x-a...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
JP Rodriguez Prof. Muralee Calc 21 28 November 2007 Written Assignment 5 Suppose a and b are positive constants, consider the function: f ( x ) = e -(( x-a)^2)/2b A&B) To find the absolute max of f ( x ), you first must find the derivative f’ ( x ) which equals f ’ ( x ) = ( e -(( x-a)^2)/2b )(2( x-a )/(2 b ) Now that we have f’ ( x ), we set it equal to zero and solve for x to find the critical numbers of the function: e -(( x-a)^2)/2b = 0 D.N.E. (-2/2 b ) x + ( a /2 b ) = 0 ( x / b ) = ( a /2 b ) x = a Now we have the critical number of f ( x ), x = a . But first we must take the second derivative and set it equal to zero to help determine where the absolute max. is. If f ’’ is negative, then the line will concave down, and if it is positive, the line will concave up. We find that: f
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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 .

Ask a homework question - tutors are online