On-line Math 21

4.2  Corollaries of the Mean Value Theorem

Theorem 1 [Generalized MVT]. If f(x) and g(x) are continuous on a closed interval [a,b] , differentiable on the open interval (a,b) , and if g¢(x) ¹ 0 for every x Î (a,b) , then there is some point c Î (a,b) so that

f¢(c) g¢(c) = f(b)-f(a) g(b)-g(a) .

Proof. This result uses a variation of the same trick as for the MVT. Construct from f and g another function, one which satisfies Rolle's theorem, and see what it tells us. Set

h(x): = f(x)-f(a)- f(b)-f(a) g(b)-g(a) (g(x)-g(a)),

which is well-defined (that is, the definition makes sense, in this case meaning that the denominator is not 0) since g(a) ¹ g(b) , because (by Rolle's theorem) g¢(x) would have to be 0 somewhere in between were g(b) = g(a) .

Then, because f and g are continuous on the closed interval [a,b] , and differentiable on the open interval (a,b) , so is h(x) . Also, h(a) = h(b) = 0 , so we satisfy Rolle's theorem's hypothesis. That means that there is some c Î (a,b) so that h¢(c) = 0 . But,

h¢(x) = f¢(x)- f(b)-f(a) g(b)-g(a) g¢(x),

so

0 = h¢(c) = f¢(c)- f(b)-f(a) g(b)-g(a) g¢(c)Þ f¢(c) g¢(c) = f(b)-f(a) g(b)-g(a) ,

where that last division makes sense because g¢(x) is never 0.

Copyright (c) 2000 by David L. Johnson.