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Unformatted text preview: Differentiation Written by MenGen Tsai email: b89902089@ntu.edu.tw 1. Let f be defined for all real x , and suppose that  f ( x ) f ( y )  ≤ ( x y ) 2 for all real x and y . Prove that f is constant. Proof:  f ( x ) f ( y )  ≤ ( x y ) 2 for all real x and y . Fix y ,  f ( x ) f ( y ) x y  ≤  x y  . Let x → y , therefore, ≤ lim x → y f ( x ) f ( y ) x y ≤ lim x → y  x y  = 0 It implies that ( f ( x ) f ( y )) / ( x y ) → 0 as x → y . Hence f ( y ) = 0, f = const . 2. Suppose f ( 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 differentible, and that g ( f ( x )) = 1 f ( x ) ( a < x < b ) . Proof: For every pair x > y in ( a, b ), f ( x ) f ( y ) = f ( c )( x y ) where y < c < x by MeanValue Theorem. Note that c ∈ ( a, b ) and f ( x ) > in ( a, b ), hence f ( c ) > 0. f ( x ) f ( y ) > 0, f ( x ) > f ( y ) if x > y , f is strictly increasing in ( a, b ). Let Δ g = g ( x + h ) g ( x ). Note that x = f ( g ( x )), and thus, ( x + h ) x = f ( g ( x + h )) f ( g ( x )) , 1 h = f ( g ( x ) + Δ g ) f ( g ( x )) = f ( g + Δ g ) f ( g ) . Thus we apply the fundamental lemma of differentiation, h = [ f ( g ) + η (Δ g )]Δ g, 1 f ( g ) + η (Δ g ) = Δ g h Note that f ( g ( x )) > 0 for all x ∈ ( a, b ) and η (Δ g ) → 0 as h → 0, thus, lim h → Δ g/h = lim h → 1 f ( g ) + η (Δ g ) = 1 f ( g ( x )) . Thus g ( x ) = 1 f ( g ( x )) , g ( f ( x )) = 1 f ( x ) . 3. Suppose g is a real function on R 1 , with bounded derivative (say  g  ≤ M ). Fix > 0, and define f ( x ) = x + g ( x ). Prove that f is onetoone if is small enough. (A set of admissible values of can be determined which depends only on M .) Proof: For every x < y , and x, y ∈ R , we will show that f ( x ) 6 = f ( y ). By using MeanValue Theorem: g ( x ) g ( y ) = g ( c )( x y ) where x < c < y, ( x y ) + (( x ) g ( y )) = ( g ( c ) + 1)( x y ) , that is, f ( x ) f ( y ) = ( g ( c ) + 1)( x y ) . ( * ) Since  g ( x )  ≤ M , M ≤ g ( x ) ≤ M for all x ∈ R . Thus 1 M ≤ g ( c ) + 1 ≤ 1 + M , where x < c < y . Take c = 1 2 M , and g ( c ) + 1 > where x < c < y for all x, y . Take into equation (*), and f ( x ) f ( y ) < since x y < 0, that is, f ( x ) 6 = f ( y ), that is, f is onetoone (injective). 2 4. If C + C 1 2 + ... + C n 1 n + C n n + 1 = 0 , where C , ..., C n are real constants, prove that the equation C + C 1 x + ... + C n 1 x n 1 + C n x n = 0 has at least one real root between 0 and 1. Proof: Let f ( x ) = C x + ... + C n n +1 x n +1 . f is differentiable in R 1 and f (0) = f (1) = 0. Thus, f (1) f (0) = f ( c ) where c ∈ (0 , 1) by MeanValue Theorem. Note that f ( x ) = C + C 1 x + ... + C n 1 x n 1 + C n x n ....
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This note was uploaded on 11/07/2009 for the course MATH 410 taught by Professor Ddd during the Spring '09 term at Punjab Engineering College.
 Spring '09
 ddd

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