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Unformatted text preview: Homework 1 β Math 118B, Winter 2010 Due on Thursday, January 14, 2010 1. Let f be defined for all real x , and suppose that β M > 0 and β Η« > such that  f ( x ) β f ( y )  β€ M  x β y  1+ Η« , β x, y β R . Prove that f is constant. Proof: If we divide by  x β y  , we get vextendsingle vextendsingle vextendsingle vextendsingle f ( x ) β f ( y ) x β y vextendsingle vextendsingle vextendsingle vextendsingle β€ M  x β y  Η« , β x negationslash = y β R . (1) Taking limits are x β y , we get  f β² ( y )  β€ β y β R , which implies that f β² = 0, and therefore f is constant. 2. Suppose that f β² ( x ) > 0 in ( a, b ). Prove that f is strictly increasing in ( a, b ), and let g be its inverse function. Prove that g is differentiable, and that g β² ( f ( x )) = 1 f β² ( x ) β x β ( a, b ) . Proof: Take x, y β ( a, b ), with x > y . Then β c β ( x, y ) such that f ( x ) β f ( y ) = f β² ( c )( x β y ). Since f β² ( c ) > 0 and x β y > 0, clearly f ( x ) β f ( y ) > 0, hence f ( x ) > f ( y ). Since x and y were arbitrary it follows that f is strictly increasing. As a consequence, f has an inverse, g , which is also strictly increasing, and continuous (note that f : [ a, b ] β [ f ( a ) , f ( b )] is continuous, and both [ a, b ] and [ f ( a ) , f ( b )] are compact). Since f β² ( x ) > 0, if ΞΎ negationslash = x then f ( ΞΎ ) negationslash = f ( x ) and we know that lim ΞΎ β x ΞΎ β x f ( ΞΎ ) β f ( x ) = 1 f β² ( x ) . Therefore, given Η« > 0, β Ξ΄ > 0 such that vextendsingle vextendsingle vextendsingle vextendsingle ΞΎ β x f ( ΞΎ ) β f ( x ) β 1 f β² ( x ) vextendsingle vextendsingle vextendsingle vextendsingle < Η« 1 2 if  ΞΎ β f ( x )  < Ξ΄ . Given such Ξ΄ > 0, β Ξ· > 0 such that  g ( t ) β x  < Ξ΄ if  t β f ( x )  < Ξ· . Note also that g ( f (...
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 Fall '09
 Garcia
 Math, Calculus, Mean Value Theorem, lim g, Xn

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