Let `f(x)a n dg(x)` be differentiable for `0lt=xlt=2` such that `f(0)=2,g(0)=1,a n df(2)=8.` Let there exist a real number `c` in `[0,2]` such that `f^(prime)(c)=3g^(prime)(c)dot` Then find the value of `g(2)dot`

Let f(x) and g(x) be differentiable for 0<=x<=1 such that f(0)=0 g(0)=6 .let there exists a real number c in (0 1) such that f'(c)=2g'(c) then the value of g(1) must be

Let f(x) and g(x) be differentiable for 0<=x<=1, such that f(0)=0,g(0)=0,f(1)=6. Let there exists real number c in (0,1) such taht f'(c)=2g'(c)* Then the value of g(1) must be 1( b) 3(c)-2(d)-1

Let f(x) and g(x) be the differentiable functions for 1<=x<=3 such that f(1)=2=g(1) and f(3)=10 .Let there exist exactly one real number c in(1,3) such that 3f'(c)=g'(c) ,then the value of g(3) must be

f(x) and g(x) are differentiable functions for 0 le x le 2 such that f(0)=5, g(0)=0, f(2)=8, g(2)=1 . Show that there exists a number c satisfying 0 lt c lt 2 and f'(c)=3g'(c) .

f(x) and g(x) are differentiable functions for 0 <= x <= 2 such that f(0) = 5, g(0) = 0, f(2)= 8,g(2) = 1. Show that there exists a number c satisfying 0 < c < 2 and f'(c)=3 g'(c).

Let f and g be differentiable on [0,1] such that f(0)=2,g(0),f(1)=6 and g(1)=2. Show that there exists c in(0,1) such that f'(c)=2g'(c)

If f is a continuous function on [0,1], differentiable in (0, 1) such that f(1)=0, then there exists some c in (0,1) such that cf^(prime)(c)-f(c)=0 cf^(prime)(c)+cf(c)=0 f^(prime)(c)-cf(c)=0 cf^(prime)(c)+f(c)=0

If f: RvecR is a differentiable function such that f^(prime)(x)>2f(x)fora l lxRa n df(0)=1,t h e n : f(x) is decreasing in (0,oo) f^(prime)(x) e^(2x)in(0,oo)

Statement 1: If both functions f(t)a n dg(t) are continuous on the closed interval [1,b], differentiable on the open interval (a,b) and g^(prime)(t) is not zero on that open interval, then there exists some c in (a , b) such that (f^(prime)(c))/(g^(prime)(c))=(f(b)-f(a))/(g(b)-g(a)) Statement 2: If f(t)a n dg(t) are continuou and differentiable in [a, b], then there exists some c in (a,b) such that f^(prime)(c)=(f(b)-f(a))/(b-a)a n dg^(prime)(c)(g(b)-g(a))/(b-a) from Lagranes mean value theorem.