Let f and g be function continuous in [a,b] and differentiable on [a,b] If f(a)=f(b)=0 then show that there is a point c in(a,b) such that g'(c)f(c)+f'(c)=0

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.

If f(x) and g(x) are continuous functions in [a,b] and are differentiable in (a,b) then prove that there exists at least one c in(a,b) for which.det[[f(a),f(b)g(a),g(b)]]=(b-a)det[[f(a),g'(c)]], where a

Let f(x)be continuous on [a,b], differentiable in (a,b) and f(x)ne0"for all"x in[a,b]. Then prove that there exists one c in(a,b)"such that"(f'(c))/(f(c))=(1)/(a-c)+(1)/(b-c).

If f is a continuous function on the interval [a,b] and there exists some c in(a,b) then prove that int_(a)^(b)f(x)dx=f(c)(b-a)

If f(x) and g(x) are continuous and differentiable functions, then prove that there exists c in [a,b] such that (f'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1.

If the function f(x) and g(x) are continuous in [a, b] and differentiable in (a, b), then the f(a) f (b) equation |(f(a),f(b)),(g(a),g(b))|=(b-a)|(f(a),f'(x)),(g(a),g'(x))| has, in the interval [a,b] :

Let f(x) and g(x) be differentiable function in (a,b), continuous at a and b, and g(x)!=0 in [a,b]. Then prove that (g(a)f(b)-f(a)g(b))/(g(c)f'(c)-f(c)g'(c))=((b-a)g(a)g(b))/((g(c))^(2))

If f(x)=(ax^(2)+b)^(3), then find the function g such that f(g(x))=g(f(x))