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

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)

Let 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 [a,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 continuous 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 Lagranges mean value theorem.

f: R to R is a continuous and f(x)=int_(0)^(x)f(t)dt then f(log_(e)5)= ?

Let us consider a function f : R rarrR , defined by f(x) =x^3 - 6 . Show that the mapping f is bijective.

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 e a

A function f : R to R is defined by f(x) =x^(2) .Then show that f is neither nor surfection .

Let f be any continuously differentiable function on [a,b] and twice differentiable on (a,b) such that f(a)=f'(a)=0 . Then

Let f be any continuously differentiable function on [a, b] and twice differentiable on (a, b) such that f(a)=f'(a)=0" and "f(b)=0 . Then-