Question

supposing that f,g:[a,b]→R are continuous functions differentiable on (a,b)(a,b) and h(t)=(f(b)−f(a))g(t)−(g(b)−g(a))f(t)h(t)=(f(b)−f(a))g(t)−(g(b)−g(a))f(t), there is some t∈(a,b)t∈(a,b) with h′(t)=0h′(t)=0

Answer 1

Answer:

See proof below

Step-by-step explanation:

This is known as the generalized mean value theorem. To prove it, the idea is to apply the mean value theorem to an adequate function.

Define the function $s(x)=\frac{1}{b-a}[f(x)(g(b)-g(a))-g(x)(f(b)-f(a))]$. Because f and g are differentiable on (a,b) and continuous on [a,b], s is also differentiable on (a,b) and continuous on [a,b]. Therefore, applying the mean value theorem to s on [a,b], there exist some t∈(a,b) such that (b-a)s'(t)=s(b)-s(a).

From the definition of s, we see that s(b)=s(a)=0 then s(b)-s(a)=0. Furthermore, using the definions of s and h, (b-a)s'(t)=-(f(b)−f(a))g('t)+(g(b)−g(a))f'(t)=-h'(t). From this, we conclude that -h'(t)=0, then h'(t)=0 and t is the required point.