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Unformatted text preview: Physics 2 Quiz 4 Solutions 1. Consider the function f ( x ) = ( x 1) exp ( 2 x 2 ). (a) Find the mimimum value of this function. (b) Find the function of the tangent to this function at x = 1. (a) To find the minimum, we first find the critical points by setting the derivative of f ( x ) equal to zero. The x crit s that are found can be reinserted into f ( x ) to identify the minimum. f ( x ) = ( x 1) e 2 x 2 (1) f ( x ) = e 2 x 2 + ( x 1)( 4 x ) e 2 x 2 (2) 0 = e 2 x 2 + ( x 1)( 4 x ) e 2 x 2 (3) 0 = (1 + ( x 1)( 4 x )) e 2 x 2 (4) 0 = ( 4 x 2 + 4 x + 1) e 2 x 2 (5) 0 = ( x 1 2 2 )( x + 1 2 2 ) e 2 x 2 (6) x crit = { 1 2 2 } (7) f ( 1 + 2 2 ) = . 01123 (8) f ( 1 2 2 ) = 1 . 108 (9) x min = 1 2 2 (10) (b) A tangent line has the form y ( x ) = mx + b where we set m = f ( x = 1) . To find b, we set y ( x = 1) = f ( x = 1) and solve. f ( x ) = ( x 1) e 2 x 2 (11) f ( x ) = ( 4 x 2 + 4 x + 1) e 2 x 2 (12) m = f (1) = e 2 (13) mx + b...
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This note was uploaded on 11/21/2010 for the course PHYSICS Physics 2 taught by Professor Manojkaplinghat during the Fall '10 term at UC Irvine.
 Fall '10
 ManojKaplinghat
 Physics

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