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.

the workers who do no drive has no reasonable reason to buy gasoline due to they don't have anything to use it for

irrational, then, just means all the numbers that aren't rational. every integer is a rational number, since each integer n can be written in the form n/1. however, numbers like 1/2, 45454737/2424242, and -3/7 are also rational, since they are fractions whose numerator and denominator are integers.

step-by-step explanation: