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`

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 be 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

If f be 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

Statement 1: If 27 a+9b+3c+d=0, then the equation f(x)=4a x^3+3b x^2+2c x+d=0 has at least one real root lying between (0,3)dot Statement 2: If f(x) is continuous in [a,b], derivable in (a , b) such that f(a)=f(b), then there exists at least one point c in (a , b) such that f^(prime)(c)=0.

A function f is differentiable in the interval 0

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

Let f(x), be a function which is continuous in closed interval [0,1] and differentiable in open interval 10,1[ Show that EE a point c in]0,1[ such that f'(c)=f(1)-f(0)

If f is continuous and differentiable function and f(0)=1,f(1)=2, then prove that there exists at least one c in[0,1] for which f'(c)(f(c))^(n-1)>sqrt(2^(n-1)), where n in N.

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