Cauchys mean value theorem suppose that f and g are

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Cauchy’s Mean Value Theorem. Suppose that f and g are continuous on [ a, b ] and differentiable on ( a, b ). Suppose further that g 0 is never zero on ( a, b ). Then there is some c ( a, b ) such that f ( a ) - f ( b ) g ( a ) - g ( b ) = f 0 ( c ) g 0 ( c ) . Proof. Note first of all that g ( a ) - g ( b ) 6 = 0. Indeed, if g ( a ) = g ( b ) then, by Rolle’s theorem, g 0 ( x ) = 0 at some point x ( a, b ). Let ϕ ( x ) = ( g ( b ) - g ( a )) f ( x ) - ( f ( b ) - f ( a )) g ( x ) + f ( b ) g ( a ) - f ( a ) g ( b ) . Then ϕ ( a ) = ϕ ( b ) = 0 and ϕ satisfies the conditions of Rolle’s theorem. Thus there exists c ( a, b ) such that ϕ 0 ( c ) = ( g ( b ) - g ( a )) f 0 ( c ) - ( f ( b ) - f ( a )) g 0 ( c ) = 0 . This implies the required result. 2
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