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Unformatted text preview: Then: there exists c such that a<c<b and f(c)=f(b)f(a)/ba Proof: let: y(x)=f(a)+(f(b)f(a)/ba))(xa) H(x)= f(x)y(x) 1. h is continuous on [a,b] 2. h is differentiable on [a,b] 3. h(a)=f(a)y(a)=0 4. h(b)=f(b)y(b)=0 5. h(a)=h(b) There exists c such that a<c<b and h(c)=0 (Rolles) H(c)=f(c)y(c)=0 = f(c)[f(b)f(a)/ba]=0 F(c)=[f(b)f(a)/ba]=0...
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This note was uploaded on 04/07/2008 for the course MATH 161 taught by Professor Algier during the Spring '08 term at Grove City.
 Spring '08
 Algier
 Mean Value Theorem

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