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

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 . 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.

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.