Suppose `f` is a continuous map from `R` to `R` and `f(f(a))=a` for some `adot` Show that there is some `b` such that `f(b)=bdot`

Let f be a continuous function defined onto [0,1] with range [0,1]. Show that there is some c in [0,1] such that f(c)=1-c

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 .

If f(x) is continuous in [a , b] and differentiable in (a,b), then prove that there exists at least one c in (a , b) such that (f^(prime)(c))/(3c^2)=(f(b)-f(a))/(b^3-a^3)

Leg f be continuous on the interval [0,1] to R such that f(0)=f(1)dot Prove that there exists a point c in [(0,1)/2] such that f(c)=f(c+1/2)dot

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) exists for all points in [a,b] and (f(c )-f(a))/(c-a)=(f(b)-f( c))/(b-c),"where"a lt clt b, then show that there exists a number 'k' such that f"(k)=0.

Let f be a continuous function on [a , b]dot If F(x)=(int_a^xf(t)dt-int_x^bf(t)dt)(2x-(a+b)), then prove that there exist some c in (a , b) such that int_a^cf(t)dt-int_c^bf(t)dt=f(c)(a+b-2c)dot

If f(x)a n dg(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. |f(a)f(b)g(a)g(b)|=(b-a)|f(a)f^(prime)(c)g(a)g^(prime)(c)|,w h e r ea

Let f:R->R be a continuous function such that |f(x)-f(y)|>=|x-y| for all x,y in R ,then f(x) will be

If a continuous function f defined on the real line R assume positive and negative values in R, then the equation f(x)=0 has a root in R. For example, if it is known that a continuous function f on R is positive at some point and its minimum value is negative, then the equation f(x)=0 has a root in R. Consider f(x)= ke^(x)-x , for all real x where k is a real constant. The line y=x meets y=ke^(x) for k le 0 at