76 CAUCHY'S PROOF OF TAYLOR'S THEOREM. 



F(a+h)-F(a) F'(a+6h) 

 f(a + h)-f(a) f'(a+eh)' 



Let s _ 



~ 



then since f'(x) is continuous and does not vanish between 

 the values 'a and a + h of x, it retains the same sign ; and 

 thus f (x) continually increases or continually decreases: see 

 Art. 89. Hence f(a + Ji) f(a) cannot be zero, and we may 

 therefore multiply by it ; so that 



F(a + A) - F(a) - E {/(a + h) -/(a)} = 0. 

 Let $ (x) denote the function 



F (a + h) - F(x) - R [f(a + h) -f(x}} : 



then <j> (x) is continuous while x lies between a and a + h ; 

 and so also is the differential coefficient <j>'(x}, that is 

 F(x) + Rf'(x}. Moreover </> (x) vanishes, by hypothesis, 

 when x = a ; and <f> (x} obviously vanishes when x = a+h. 

 Hence, by Art. 91, it follows that </>' (x} must vanish for some 

 value of x between a and a+ k; this value may be denoted 

 by a + Oh, where 6 is some proper fraction. Thus 



-F(a + eh] + Bf'(a + 6h) = ; 



and, by hypothesis, /' (a + Oh] is not zero, so that we may 

 divide by it : therefore 



F'(a+0h) 



Thus the required result is obtained. 



99. The result of the preceding Article has been obtained 

 on the assumption that the functions are continuous and that 

 f (x} does not vanish between the values a and a + h of the 

 variable x. The result however is true if the functions are 

 continuous and either of the two F' (x) and f'(x) does not 

 vanish. For if F' (x) does not vanish we may prove as in 

 the preceding Article that 



f(a) f'(a+0h) 



